Inequalities

Last updated Jul 5, 2023

Inequalities are types of questions which involve two algebraic expressions, one of which is lesser than the other. In school maths, inequality questions generally ask you to find a range of values that satisfy the inequality. For example, I’m sure most of you have seen something similar to
’Find all x so that x + 3 ≥ 12’. The other class of inequalities is the main one that appears in olympiads, and the one this handout focuses on. These types of questions provide you with two algebraic expressions, and require that you prove that one is <,≤,≥, or > than the other.

You should probably know how to do this before moving on.

• a > b and c ≥ d ⟹ a + c > b + d

• If c > 0, then a > b ⟹ ac > bc

• And if c < 0, then a > b ⟹ ac < bc

• For m > 0 and a > b, a^m > b^m. If m < 0, then a^m < b^m

The trivial inequality: x² ≥ 0 for any real x, with equality at x = 0.
This inequality is the foundation of several more, and something you must know for any olympiad. It seems incredibly obvious, but it is also quite powerful and can be used to derive a variety of results. For example, try the following problem only using the trivial inequality: Prove that for all triplets (a,b,c),  we have a² + b² + c² ≥ ab + bc + ac The more astute among you might immediately recognise what to do here. If you can’t see it yet: don’t worry. It’s not very obvious to people who are new to inequalities.
We already know that we should be using the trivial inequality in some way, so how?
Well, let’s see what happens when we rearrange it: we have a² + b² + c² − ab − ac − bc. It may remind you of the expansion of (a−b)², but it’s off by a factor of 2. So, consider the following 2a² + 2b² + 2c² − 2ab − 2ac − 2bc, which upon rearranging yields a² − 2ab + b² + a² − 2ac + c² + b² − 2bc + c² = (a−b)² + (a−c)² + (b−c)². By the trivial inequality, all of these squares must be at least 0, so (a−b)² + (a−c)² + (b−c)² ≥ 0. Upon dividing this by two, we get the original statement.

We can now use the trivial inequality to prove a much more powerful inequality.

$$\text{For any positive reals } a_1, a_2, …, a_n \text{ we have } \frac{a_1 + a_2 + … + a_n}{n} \geq \sqrt[n]{a_1a_2\cdots a_n}$$ Ok, I lied a bit. While we can use the trivial inequality to prove it for all n, the process is rather time-consuming, complicated, and just generally annoying to write. We can, however, prove it fairly easy for n = 2. $$\text{Prove that, for any two positive reals } a, b, \hspace{10pt} \frac{a+b}{2} \geq \sqrt{ab}$$ This looks weirdly close to a perfect square, except for that annoying denominator on the LHS. So, we can double both sides and rearrange to get: $$a - 2\sqrt{ab} + b \geq 0$$ Which is indeed the expansion of $(\sqrt{a} - \sqrt{b})^2$, and so by the trivial inequality we are done.

Let’s try a simple inequality with the power of AM-GM: Let a, b, c ≥ 0. Then, prove that (a+b)(a+c)(b+c) ≥ 8abc We can immediately see that we want to prove that a product of sums is less than a product. In most of these kinds of problems, AM-GM will do you just fine. So, we take a leap of faith and apply AM-GM to each of the bracketed pairs (I divided each term by 2 to make it clearer): $$(a+b)(a+c)(b+c) = 8(\frac{a+b}{2})(\frac{a+c}{2})(\frac{b+c}{2}) \geq 8(\sqrt{ab})(\sqrt{ac})(\sqrt{bc}) = 8 \sqrt{a^2b^2c^2} = 8abc.$$ And we’re done. The AM-GM inequality is actually a corollary of a much stronger, generalised inequality.

Warning: Yes, what you are about to see is very complicated. You don’t need to memorise this for AMO or ASC - only the most important corollaries are required, which I will discuss after.

$$\text{For some positive reals } a_1, a_2,…,a_n, \text{ define}$$

$$M_r = (\frac{a_1^r+a_2^r+…+a_n^r}{n})^{\frac{1}{r}} \text{ for all non-zero reals } r \text{ and } M_0 = \sqrt[n]{a_1a_2…a_n}.$$

Then, for r > s, M_r ≥ M_s with equality if and only if a₁ = a₂ = … = a_n

You might notice that M₁ is actually just the AM, and M₀ is just the GM. So, we already have the generalised version of AM-GM just from this. But, there are only 4 means that are actually relevant.
The actually important means: Quadratic - Arithmetic - Geometric - Harmonic

$$\text{Define } QM = M_2, AM = M_1, GM = M_0, HM = M_{-1}. $$

$$\text{Then, } M_2 \geq M_1 \geq M_0 \geq M_{-1} \text{ or, more clearly, }$$

$$\sqrt{\frac{a_1^2 + a_2^2 + … + a_n^2}{n}} \geq \frac{a_1 + a_2 + … + a_n}{n} \geq \sqrt[n]{a_1a_2…a_n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + … + \frac{1}{a_n}}$$

It looks scary, but it’s actually a really powerful tool. Before we move on, see if you can figure out how to prove GM-HM (i.e. the rightmost part)! Hint: you’ll have to assume AM-GM is true and use that (I can’t be bothered proving AM-GM for all n here, it’s far too long for me to care. Just believe that it’s true and it will be - a surprisingly useful strategy in olympiads)

Now, let’s try tackle some problems using the powerful power means inequality. $$\text{Suppose } x_1, x_2 … x_n \text{ are positive reals so that } x_1x_2 \cdots x_n = 1.$$

$$\text{Then, for } r > s > 0, \text{ we have } x_1^r + x_2^r + … + x_n^r \geq x_1^s + x_2^s + … + x_n^s$$

This is about as close to power means as you can get, specifically M_r and M_s. Writing this out:

$$(\frac{x_1^r+x_2^r+…+x_n^r}{n})^{\frac{1}{r}} \geq (\frac{x_1^s+x_2^s+…+x_n^s}{n})^{\frac{1}{s}}$$ Which, upon raising both sides to the power of r, rearranges to $$\frac{x_1^r + x_2^r … + x_n^r}{n} \geq (\frac{x_1^s+x_2^s+…+x_n^s}{n})^{\frac{r}{s}} = (\frac{x_1^s+x_2^s+…+x_n^s}{n}) \cdot (\frac{x_1^s+x_2^s+…+x_n^s}{n})^{\frac{r}{s}-1}$$ But, by assumption we have $\frac{r}{s} - 1 > 0$, and by AM-GM we have $$\frac{x_1^s + x_2^s + … + x_n^s}{n} \geq \sqrt[n]{x_1^sx_2^s…x_n^s} = 1 \implies (\frac{x_1^s + x_2^s + … + x_n^s}{n})^{\frac{r}{s} - 1} \geq 1$$ $$\text{So, } \frac{x_1^r + x_2^r + … + x_n^r}{n} \geq (\frac{x_1^s+x_2^s+…+x_n^s}{n}) \cdot (\frac{x_1^s+x_2^s+…+x_n^s}{n})^{\frac{r}{s}-1} \geq \frac{x_1^s+x_2^s+…+x_n^s}{n}$$ And multiplying by n yields the desired inequality. This inequality is also quite useful in it’s own right, so it’s definitely worth keeping in mind.

$$\text{For two sequences of real numbers } \{ x_1, x_2, …, x_n\} \text{ and } \{ y_1, y_2, …, y_n\}, \text{ we have }$$ $$(x_1^2 + x_2^2 + … + x_n^2)(y_1^2 + y_2^2 + … + y_n^2) \geq (x_1y_1 + x_2y_2 + … + x_ny_n)^2 $$ with equality if and only if there exists some constant c such that y_i = cx_i for all i. Cauchy-Schwarz is quite a powerful inequality, especially when it comes to dealing with squares or square roots. As the holy grail of mathematics, Problem Solving Tactics, says: $$\textit{If there are squares or square roots, try Cauchy-Schwarz.}$$ $$\textit{If there are products or fractions, try Cauchy-Schwarz.}$$ If all else fails, try Cauchy-Schwarz. So, with the power of Cauchy-Schwarz on our side, let’s try solve an problem from a while back: $$\text{Prove AM-HM, that is, } \frac{x_1 + x_2 + … + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + … + \frac{1}{x_n}}$$ Our first step when looking at this should probably be to recoil in fear at the fractions-in-a-fraction. So, rearranging both sides allows us to rephrase the problem statement as: $$(x_1 + x_2 + … + x_n)(\frac{1}{x_1} + \frac{1}{x_2} + … + \frac{1}{x_n}) \geq n^2$$ Which is very Cauchy looking. Indeed, applying Cauchy-Schwarz to the sequences $\{\sqrt{x_1}, \sqrt{x_2}, …, \sqrt{x_n}\}$ and $\{ \frac{1}{\sqrt{x_1}}, \frac{1}{\sqrt{x_2}}, …, \frac{1}{\sqrt{x_n}}\}$ yields $$(x_1 + x_2 + … + x_n)(\frac{1}{x_1} + \frac{1}{x_2} + … + \frac{1}{x_n}) \geq (\frac{\sqrt{x_1}}{\sqrt{x_1}} + \frac{\sqrt{x_2}}{\sqrt{x_2}} + … + \frac{\sqrt{x_n}}{\sqrt{x_n}})^2 = n^2$$

$$\text{If } a_1 \geq a_2 \geq … \geq a_n \text{ and } b_1 \geq b_2 \geq … \geq b_n \text{ are any two sequences of reals,}$$ then for any permutation β₁, β₂, …, β_n of b₁, b₂, …, b_n we have the following inequality: a₁b₂ + a₂b₂ + … + a_nb_n ≥ a₁β₁ + a₂β₂ + … + a_nβ_n ≥ a₁b_n + a₂b_n − 1 + … + a_nb₁

It looks a bit complicated, so you can essentially think of it as saying that the product of two sequences is maximised when the biggest ones are paired up, and minimised when the biggest and smallest ones are paired up. Let’s try a problem using this. For positive reals, prove that a^ab^bc^c ≥ a^bb^cc^a Handy trick for these kinds of problems: If there’s a lot of weird powers, try logs!. Since the function log(x) is increasing, we can take the logarithm of both sides without changing the inequality: alog(a) + blog(b) + clog(c) ≥ alog(b) + blog(c) + clog(a) (Simplified using log laws). But upon considering the two sequences a, b, c and log(a), log(b), log(c), and noting that a ≥ b ⟹ log(a) ≥ log(b), we find that, by the rearrangement inequality, their products are maximised when sorted the same way, easily solving the question.

The inequalities in this section are weirder, and while they are helpful in higher level olympiads (AMO Q4/8’s, EGMO, IMO), you’ll almost never need to use them. Also, they’re sort of fringe inequalities, so I haven’t explained or provided example problems here, only their statement.

Jensen’s inequality If the real numbers x₁, x₂, …, x_n lie on an interval where some function f(x) is convex, then $$\frac{f(x_1) + f(x_2) + … + f(x_n)}{n} \geq f(\frac{x_1 + x_2 + … + x_n}{n})$$ $$\text{and the inequality is reversed if $f(x)$ is concave on that interval.}$$ Weighted Power Means $$\text{Given positive reals } x_1, x_2, …, x_n \text{ and weights } w_1, w_2, …, w_n, \text{ with sum $w$, define $M_r$ as }$$ $$M_r = (\frac{w_1a_1^r+w_2a_2^r+…+w_na_n^r}{w})^{\frac{1}{r}} \text{ for all non-zero reals } r \text{ and } M_0 = \sqrt[w]{a_1^{w_1}a_2^{w_2}…a_n^{w_n}}.$$ Then, M_r ≥ M_s for r > s, with equality if and only if a₁ = a₂ = … = a_n or w₁ = w₂ = … = w_n = 0 $$\text{Setting } w_1 = w_2 = … = w_n = 1 \text{ results in the original unweighted power means inequality.}$$ Hölder’s inequality $$\text{Let } a_1, a_2, …, a_n \text{ and } b_1, b_2, …, b_n \text{ be two sets of non-negative reals. Let $p, q \in \mathbb{R}$ such that } \frac{1}{p} + \frac{1}{q} = 1.$$ $$\text{Then, } (a_1^p + a_2^p + … + a_n^p)^{\frac{1}{p}}(b_1^q + b_2^q + … + b_n^q)^{\frac{1}{q}} \geq a_1b_1 + a_2b_2 + … + a_nb_n $$ Tangent-line trick: $$\text{Suppose we have reals } x_1, x_2, …, x_n \text{ that sum to $s$. Then, for some function}$$ $$f(x) \geq ax + b \text{ for all $x$ and some constants $a,b$, we have } f(x_1) + f(x_2) + … f(x_n) \geq as + bn.$$ Muirhead’s inequality: Very cool inequality but it will literally never be useful. $$\text{A sequences of reals } A = (a_1, a_2, …, a_n) \text{ is said to majorise another sequence } B = (b_1, b_2, …, b_n) \text{ if} $$ $$a_1 \geq a_2 \geq … \geq a_n \text{ and } b_1 \geq b_2 \geq … \geq b_n \text{ and for all } 1 \leq i < n, \text{ we have} $$ a₁ + a₂ + … + a_i ≥ b₁ + b₂ + … + b_i,  along with a₁ + a₂ + … + a_n = b₁ + b₂ + … + b_n. So, Muirhead’s inequality says that if A majorises B, then we have: $$\sum_{sym} x_1^{a_1}x_2^{a_2} \cdots x_n^{a_n} \geq \sum_{sym} x_1^{b_1}x_2^{b_2} \cdots x_n^{b_n} \hspace{10pt} \text{ for any non-negative reals } x_1, x_2, …, x_n. $$ Schur’s inequality: Also cool but probably useless. For all non-negative reals a, b, c, r,  we have: a^r(a−b)(a−c) + b^r(b−a)(b−c) + c^r(c−a)(c−b) ≥ 0

Arranged in roughly increasing difficulty

Problem 1: Find the minimum value of abc given a + 4b + 16c = 256

Problem 2: Suppose x₁, x₂, …, x_n are positive reals with product 1. Prove that (1+x₁)(1+x₂)…(1+x_n) ≥ 2ⁿ

Problem 3: If a, b, c are positive reals, prove that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a + b} \geq \frac{3}{2}$ in three different ways

Problem 4: For positive a, b, if a + b = 1, prove that $$\frac{(a+1)(b+1)}{ab} \geq \frac{9}{4}$$

Problem 5: For positive reals a, b, c, d, prove that $$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \geq \frac{64}{a+b+c+d}$$

Problem 6: Let A, B, C be the angles of an acute triangle. Prove that $$\text{ sin }A + \text{ sin }B + \text{ sin }C \leq \frac{3\sqrt{3}}{2}$$

Problem 7: Let x > 1 be a real number, and n > 1 be an integer. Prove that $$1 + \frac{x-1}{nx} < \sqrt[n]{x} < 1 + \frac{x-1}{n}$$

Problem 8: For positive reals a, b, c, prove that $$\frac{a+b+c}{3} \geq \sqrt{\frac{ab + bc + ac}{3}}$$

Problem 9: For n ≥ 3, let a₂, a₃, …, a_n be positive reals such that a₂a₃…a_n = 1. Prove that (1+a₂)²(1+a₃)³…(1+a_n)ⁿ > nⁿ

Problem 10: Let a ≥ b ≥ c ≥ d > 0 be reals so that a + b + c + d = 1. Prove that (a+2b+3c+4d)a^ab^bc^cd^d < 1

Shortlist. A few of these are A6-8’s i.e. the hardest problems proposed for the IMO, and insanely difficult to solve. So don’t feel bad if you spend a month on one and can’t solve it. The last one is the legendary IMO 2021 Q2, which only had 16 solves and an average score of 0.375/7. Good luck!

Problem 11: Prove that, for all positive reals a, b, c, we have (a²+2)(b²+2)(c²+2) ≥ 9(ab+bc+ac)

Problem 12: Suppose a, b, c, d are non-negative reals with sum 4. Determine the minimal value of $$\frac{a}{b^3+4} + \frac{b}{c^3+4} + \frac{c}{d^3+4} + \frac{d}{a^3+4}$$

Problem 13: Prove that for all positive reals a, b, c, k, we have $$\frac{a}{\sqrt{a^2 + kbc}} + \frac{b}{\sqrt{b^2 + kac}} + \frac{c}{\sqrt{c^2 + kab}} \geq \frac{3}{\sqrt{1+k}}$$

Problem 14: Let a₁, a₂, …, a_n, b₁, b₂, …, b_n be non-negative real numbers. Prove that $$\sum_{i=1}^n \sum_{j=1}^n \text{min}(a_ia_j, b_ib_j) \leq \sum_{i=1}^n \sum_{j=1}^n \text{min}(a_ib_j, a_jb_i)$$

Problem 15: Let x₁, x₂, …, x_n, y₁, y₂, …, y_n ∈ [0,1] be reals such that x_i + y_i = 1 for all i ≤ n. For any positive integer m, prove (1−x₁x₂…x_n)^m + (1−y₁^m)(1−y₂^m)…(1−y_n^m) ≥ 1

Problem 16: Let a, b, c be positive reals such that a² + b² + c² + abc = 4. Prove that 0 ≤ ab + bc + ac − abc ≤ 2

Problem 17: Let a₁, a₂, …, a_n ∈ [1,2^k] for some n, k. Prove that $$\sum_{i=1}^n \frac{a_i}{\sqrt{a_1^2 + … + a_i^2}} \leq 4\sqrt{kn}$$

Problem 18: If a, b, c, d are non-negative reals with sum 100, find the maximal value of $$\sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}}$$

Problem 19: Prove that, for all x₁, x₂, …x_n ∈ ℝ $$\sum_{1 \leq i \leq j \leq n}\sqrt{|x_i - x_j|} \leq \sum_{1 \leq i \leq j \leq n}\sqrt{|x_i + x_j|} \\$$