3)
Let $a=x+y$, $b=y+z$, and $c=x+z$. The inequality becomes
$$\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}+\frac{x+z}{x+2y+z}+\frac{x^2+y^2+z^2+3(xy+yz+xz)}{2(x^2+y^2+z^2)+2(xy+yz+xz)}\le \frac52$$
$$\Longleftrightarrow$$
$$\frac12+\frac{x^2+y^2+z^2+3(xy+yz+xz)}{2(x^2+y^2+z^2)+2(xy+yz+xz)}\le 2\left[\frac{z}{x+y+2z}+\frac{x}{2x+y+z}+\frac{y}{x+2y+z}\right]$$
$$\Longleftrightarrow$$
$$\frac{(x+y+z)^2}{x^2+y^2+z^2+xy+yz+xz}\le 2\left[\frac{z}{x+y+2z}+\frac{x}{2x+y+z}+\frac{y}{x+2y+z}\right]$$
But by Cauchy-Schwartz,
$$\frac{z}{x+y+2z}+\frac{x}{2x+y+z}+\frac{y}{x+2y+z}\ge \frac{(x+y+z)^2}{2(x^2+y^2+z^2)+2(xy+yz+xz)}$$
Bài 3) Inequality is equivalent to
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq \frac{3}{2}+\frac{\sum (a-b)^2}{2(a^2+b^2+c^2)}$
Using Cauchy-Schwarz, we have
$3-(\sum \frac{a}{b+c})=\sum \frac{b+c-a}{b+c}=\sum \frac{(b+c-a)^2}{(b+c)(b+c-a)}\geq \frac{(\sum a)^2}{2(a^2+b^2+c^2)}$
$\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq \frac{3}{2}+\frac{\sum (a-b)^2}{2(a^2+b^2+c^2)}$
We are done
Second solution: Let $a=x+y;b=y+z;c=z+x$. Inequality is equivalent to
$\sum \frac{x+y}{x+y+2z}\leq \frac{3}{2}+\frac{\sum (x-y)^2}{2[\sum (x+y)^2]}$
$\sum \frac{x+y}{x+y+2z}\leq 3-\frac{2(x+y+z)^2}{\sum (x+y)^2}$
$\sum \frac{-2z}{x+y+2z}\leq \frac{-2(x+y+z)^2}{\sum (x+y)^2}$
$\sum \frac{z}{x+y+2z}\geq \frac{(x+y+z)^2}{2(x^2+y^2+z^2+xy+yz+xz)}$
The last is true by Cauchy-Schwarz
Bài 3) We can write your inequality as\[\frac{2(p^2-Rr-r^2)}{p^2+2Rr+r^2}+\frac{p^2+4Rr+r^2}{2(p^2-4Rr-r^2)} \leqslant \frac{5}{2},\]equivalent to\[(2R+3r)p^2-4Rr(8R+7r)-5r^3 \geqslant 0,\]
or
\[10r^2(R-2r)+(2R+3r)(p^2+5r^2-16Rr) \geqslant 0.\]
Which is true.
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