Prove: $\frac{x+ 2\,y}{x^{2}+ 3\,y+ 5}+ \frac{y+ 2\,x}{y^{2}+ 3\,x+ 5}+ \frac{1}{4\left ( x+ y- 1 \right )}\geqq \frac{7}{8}$

Question

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!

Answer 1

After your substitution we obtain: $$27u^4v^4+84u^3v^3(u+v)+116u^2v^2(u^2+v^2)+345u^3v^3+$$ $$+uv(64u^3+645u^2v+645uv^2+64v^3)+uv(484u^2+1408uv+484v^2)+1161uv(u+v)+$$ $$+252(u^2+v^2)+983uv+222(u+v)+45\geq0,$$ which is obvious.

Answer 2

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 $t \in [2, 4]$.