<s>[INST] Prove that $\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}$ How to prove
\begin{equation*}
\frac{1}{2}(a+b)^2+\frac{1}{4}(a+b)\ge a\sqrt{b}+b\sqrt{a}\ ,where \   a,b\ge 0  
\end{equation*}
I applied AM-GM inequality but obtained only
\begin{equation*}
\frac{3}{2}(a+b)\ge a\sqrt{b}+b\sqrt{a}+1 
\end{equation*}
which is not true for, example, $a=0, \ b=\frac{1}{3}$
 [/INST] From first principles:
$$(\sqrt{x}-\sqrt{y})^2 \geq 0$$
$$x+y-2\sqrt{xy} \geq 0$$
$$\sqrt{xy} \leq \frac{x+y}{2}$$
And:
$$(\sqrt{x}-\frac12)^2 \geq 0$$
$$x+\frac14-\sqrt{x} \geq 0$$
$$\sqrt{x} \leq x + \frac14$$
Given that:
$$\begin{split}
a\sqrt{b}+b\sqrt{a} &= \sqrt{ab}(\sqrt{a}+\sqrt{b}) \\
&\leq\frac{a+b}2(\sqrt{a}+\sqrt{b}) \\
&\leq\frac{a+b}2(a+b+\frac12) \\
&=\frac12(a+b)^2+\frac14(a+b)
\end{split}$$
$$$$
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