
6. Find the greatest real $k$ such that

$$
\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqslant \sqrt{a b}
$$

holds for any positive real $a$ and $b$.

(Ján Mazák)


Solution. If $k=2$ then the inequality is equivalent to $\frac{1}{2}(a+b) \geqslant \sqrt{a b}$ (which holds), therefore $k \geqslant 2$. For $k>2$ we have $k+2>0$

$$
2\left(a^{2}+k a b+b^{2}\right) \geqslant(k+2)(a+b) \sqrt{a b},
$$

the division by $b^{2}$ gives

$$
2\left(\frac{a^{2}}{b^{2}}+k \frac{a}{b}+1\right) \geqslant(k+2)\left(\frac{a}{b}+1\right) \sqrt{\frac{a}{b}} .
$$

Denote $\sqrt{a / b}=x$. The value of $x$ can be any positive number. Thus the problem is equivalent to find greatest $k$, such that

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

holds for any positive $x$. We get

$$
\begin{aligned}
k\left(\left(x^{2}+1\right) x-2 x^{2}\right) & \leqslant 2\left(x^{4}+1-\left(x^{2}+1\right) x\right), \\
k\left(x^{3}-2 x^{2}+x\right) & \leqslant 2\left(x^{4}-x^{3}-x+1\right), \\
k x\left(x^{2}-2 x+1\right) & \leqslant 2\left(x^{3}(x-1)-(x-1)\right), \\
k x(x-1)^{2} & \leqslant 2(x-1)^{2}\left(x^{2}+x+1\right) .
\end{aligned}
$$

For $x=1$ the inequality holds. If $x \neq 1$ then the division by positive number $x(x-1)^{2}$ gives

$$
k \leqslant \frac{2\left(x^{2}+x+1\right)}{x}=2+2\left(x+\frac{1}{x}\right) .
$$

If $x \neq 1$ then the values of $x+1 / x$ fill the whole interval $(2, \infty)$, the RHS can be any number in $(6, \infty)$, consequently $k \geqslant 6$.
