<s>[INST] Find $k_{max}$ if such $\frac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}$ Let $a,b>0$ then we have
$$\color{crimson}{\dfrac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}}$$ Find $k_{\max}$
Everything I tried has failed so far.
Here is one thing I tried, but obviously didn't work.
Consider the Special case  $a=b$
then
$$\color{crimson}{\dfrac{2(a^2+kab+b^2)}{(k+2)(a+b)}=\dfrac{2(2a^2+ka^2)}{2(k+2)a}=a=RHS}$$Thanks in advance
 [/INST] The inequality can be rewritten as $$2(a-b)^2\ge (k+2)(\sqrt{a}-\sqrt{b})^2$$ which is equivalent when $a\neq b$ to $$2(\sqrt{a}+\sqrt{b})^2\ge (k+2)\sqrt{ab}$$ or $$2(\sqrt{a}-\sqrt{b})^2\ge (k-6)\sqrt{ab}$$ This shows that the inequality always holds for $k=6$. For any $k>6$, we may take $a=1, b=1+\epsilon$ for sufficiently small $\epsilon>0$, and the inequality becomes false. So $k=6$ is best. 
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