# 2001 IMO Problems/Problem 2

## Problem

Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$.

## Solution

### Solution using Holder's

By Holder's inequality, $\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}$ Thus we need only show that $(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc$ Which is obviously true since $(a+b)(b+c)(c+a)\ge 8abc$.

### Alternate Solution using Jensen's

This inequality is homogeneous so we can assume without loss of generality $a+b+c=1$ and apply Jensen's inequality for $f(x)=\frac{1}{\sqrt{x}}$, so we get: $$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}$$ but $$1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc$$ by AM-GM, and thus the inequality is proven.

### Alternate Solution using Isolated Fudging

We claim that $$\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$$ Cross-multiplying, squaring both sides and expanding, we have $$a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc$$ After cancelling the $a^{\frac{14}{3}}$ term, we apply AM-GM to RHS and obtain $$a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc$$ as desired, completing the proof of the claim.

Similarly $\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$ and $\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}$. Summing the three inequalities, we obtain the original inequality.

### Alternate Solution using Cauchy

We want to prove $$\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1$$

Note that since this inequality is homogenous, assume $a+b+c=3$.

By Cauchy, $$\left(\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\right)\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)\ge (a+b+c)^2=9$$

Dividing both sides by $\sum_{cyc}a\sqrt{a^2+8bc}$, we see that we want to prove $$\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1$$ or equivalently $$\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9$$

Squaring both sides, we have $$\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81$$

Now use Cauchy again to obtain $$\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le (a+b+c)\left(\sum_{cyc}a(a^2+8bc)\right)\le 81$$

Since $a+b+c=3$, the inequality becomes $$\sum_{cyc}a^3+8abc\le 27$$ after some simplifying.

But this equals $$(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27$$ and since $a+b+c=3$ we just want to prove $$\left(\sum_{sym}a^2b\right)\ge 6abc$$ after some simplifying.

But that is true by AM-GM or Muirhead. Thus, proved. $\Box$