<s>[INST] How prove this $xy+yz+xz\le 2xyz+\frac{1}{2}$ let $x,y,z>0$ and such
$$x^2+y^2+z^2+2xyz=1$$
show that
$$xy+yz+xz\le 2xyz+\dfrac{1}{2}$$
My try: since
$$1=x^2+y^2+z^2+2xyz\ge xy+yz+xz+2xyz$$
then
$$xy+yz+xz\le 1-2xyz$$
so
we only prove follow this 
$$1-2xyz\le 2xyz+\dfrac{1}{2}$$
$$\Longleftrightarrow xyz\ge\dfrac{1}{4}$$
But this is not true,so How prove my inequality? Thank you
 [/INST] Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ are positives.
Hence, $z=\frac{c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that
$$\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\leq\frac{2abc}{\prod\limits_{cyc}(a+b)}+\frac{1}{2}$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq4abc+\prod_{cyc}(a+b)$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}(a^2b+a^2c+2abc),$$
which is AM-GM:
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+b+2c)=\sum_{cyc}(a^2b+a^2c+2abc).$$
Done!
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