<s>[INST] Prove: $\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$ Prove: $$\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$$
for $x,\,y\in \left [ 1,\,2 \right ]$
My unsuccessful try (with my method like ABC method, but I can't continue):
We have: $$x\in \left [ 1,\,2 \right ]\rightarrow x- 1\in \left [ 0,\,1 \right ]\rightarrow \frac{1}{x- 1}\in\left [ 1, +\infty  \right )\rightarrow \underbrace{\frac{1}{x- 1}-1}_{u}\in\left [ 0, +\infty  \right )\rightarrow \underbrace{x= \frac{u+ 2}{u+ 1}}_{u>0}$$
Similarly, $\underbrace{y= \frac{v+ 2}{v+ 1}}_{v>0}$, we have a new expression:
https://www.wolframalpha.com/input/?i=(x%2B2y)%2F(x%5E2%2B3y%2B5)%2B(y%2B2x)%2F(y%5E2%2B3x%2B5)%2B1%2F(4(x%2By-1))+with+x%3D(u%2B2)%2F(u%2B1),y%3D(v%2B2)%2F(v%2B1)
That's too large & too big! I need to the help & your opinions about my method! Thanks!
 [/INST] An alternate way to your substitution:  $x\in [1, 2] \implies (x-1)(x-2)\leqslant 0 \implies x^2\leqslant 3x-2$.  Similarly $y^2\leqslant 3y-2$.  Using these in the LHS, we get
$$LHS \geqslant \frac{x+2y}{3(x+y+1)}+\frac{y+2x}{3(x+y+1)}+\frac1{4(x+y-1)}$$
Simplifying and letting $x+y=t$, we are trying to find the minimum of the univariate
$$f(t) = \frac{t}{t+1}+\frac1{4(t-1)}$$
which is easily found setting $f'(t)=0$ as $f(t) \geqslant f(3) = \frac78$ for $t \in [2, 4]$.
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