Problem 1334

## Solution to Exercise 7

In this exercise, we would like to apply the arithmetic-geometric inequality to obtain a lower bound of the form $\left(x^{2}+1\right)+\frac{1}{x^{2}+1} \geqslant 2$. However, if we approach it poorly, we cannot conclude. Indeed, we might have suspected this by studying the equality case of the equation. Given the condition $x y + y z + z x = 1$, we suspect that equality will occur for $x = y = z = \frac{1}{\sqrt{3}}$. By substituting these values into the equation, we find that equality holds, but this is not an equality case of the earlier arithmetic-geometric inequality.

In the equality case $x = y = z = \frac{1}{\sqrt{3}}$, we have $x^{2} + 1 = \frac{4}{3}$ and $\frac{1}{x^{2} + 1} = \frac{3}{4}$. We therefore apply the arithmetic-geometric inequality with appropriate coefficients to have the equality of the two terms in this case:

$$\frac{3}{4}\left(x^{2}+1\right) + \frac{4}{3}\left(\frac{1}{x^{2}+1}\right) \geqslant 2$$

By summing these three equations, we get

$$\frac{3}{4}\left(x^{2} + y^{2} + z^{2} + 3\right) + \frac{4}{3}\left(\frac{1}{x^{2}+1} + \frac{1}{y^{2}+1} + \frac{1}{z^{2}+1}\right) \geqslant 6$$

which gives

$$\frac{3}{4}\left(x^{2} + y^{2} + z^{2}\right) + \frac{4}{3}\left(\frac{1}{x^{2}+1} + \frac{1}{y^{2}+1} + \frac{1}{z^{2}+1}\right) \geqslant \frac{15}{4}$$

Compared to the inequality in the problem statement, we still need to bound the term $\frac{5}{4}\left(x^{2} + y^{2} + z^{2}\right)$. But by the arithmetic-geometric inequality,

$$\frac{5}{4}\left(x^{2} + y^{2} + z^{2}\right) \geqslant \frac{5}{4}(x y + y z + z x) = \frac{5}{4}$$

By adding this inequality to the previously obtained inequality, we get

$$2\left(x^{2} + y^{2} + z^{2}\right) + \frac{4}{3}\left(\frac{1}{x^{2}+1} + \frac{1}{y^{2}+1} + \frac{1}{z^{2}+1}\right) \geqslant \frac{15}{4} + \frac{5}{4} = 5$$

## Alternative Solution to Exercise 7

Let $s = x^{2} + y^{2} + z^{2}$. Note that by the inequality of the bad students,

$$\frac{4}{3}\left(\frac{1}{x^{2}+1} + \frac{1}{y^{2}+1} + \frac{1}{z^{2}+1}\right) \geqslant \frac{4}{3} \frac{(1+1+1)^{2}}{x^{2} + y^{2} + z^{2} + 3} = \frac{12}{s+3}$$

Thus, it suffices to show that $2s + \frac{12}{s+3} \geqslant 5$. By multiplying both sides by $s+3$ which is positive, it suffices to show that $2s(s+3) + 12 \geqslant 5(s+3)$. This is equivalent to $2s^{2} + 6s + 12 \geqslant 5s + 15$ and thus to $2s^{2} + s \geqslant 3$. By the tourniquet lemma, $s \geqslant x y + y z + x z = 1$, so $2s^{2} + s \geqslant 3$. We have the desired result.

Comment from the graders: The exercise was well solved by the students who attempted it. Be careful not to get tangled up in the inequalities. Here, we could have directly used the inequality of the bad students: setting $s = x^{2} + y^{2} + z^{2}$ allowed us to conclude by multiplying by $s+3$. Note that it is not necessary to reprove $x^{2} + y^{2} + z^{2} \geqslant x y + y z + x z$ each time to use it: this is the tourniquet lemma, which is present in the POFM course.