How to solve this simple but complicated looking inequality?

Question

Completely stuck on solving this question,

Any hints or ideas? I tried expanding but got nowhere with it, it seems to be more complicated than some $a^2\geq 0$.

Answer 1

Take the LCM on the RHS.

$\frac{2(x^2 + y^2 + xy)}{2x^2 + 2y^2 + 5 xy} \geq \frac{1}{2}$

Since we know that x and y are strictly non negative, therefore multiplying by the denominator on both sides does not change the inequality sign.

Note that x and y can't simultaneously be $0$ so we are not using 0 anywhere.

Multiply by $2(2x^2 + 2y^2 + 5 xy)$ on both sides.

$4(x^2 + y^2 + xy) \geq 2x^2 + 2y^2 + 5 xy$

Simplifying, we get

$2x^2 + 2y^2 \geq xy$

From here, we can use AM-GM on the LHS to prove that this is true for all non - negative real numbers.

$\frac{2x^2 + 2y^2}{2} \geq \sqrt(2x^2 \cdot 2y^2) $

$\frac{2x^2 + 2y^2}{2} \geq 2xy$

$2x^2 + 2y^2 \geq 4xy$

Answer 2

Well if $x,y =0$ then the left hand side would be undefined the statement is not true. But it is true for all $x, y: x\ge 0, y\ge 0$ and not both $x,y=0$.

If $x=0$ or $y=0$ then either $\frac x{x+2y} = \frac xx = 1$ or $\frac y{y+2x} =\frac yy =1$. In which case the LHS is at least $1$.

So we just need to show for the cases when $x\ne 0$ and $y \ne 0$. But then both $x+2y$ and $y+2x$ are positive and

$\frac x{x+2y} + \frac y{y+2x} \ge \frac 12$ if and only if

$2x(y+2x) + 2y(x+2y) \ge (x+2y)(y+2x)$ if and only if

$4x^2 + 4xy + 4y^2 \ge 2x^2 + 5xy + 2y^2$ if and only if

$2x^2 + 2y^2 \ge xy$.

That's obviously true.

The AM-GM inequality says $x^2 + y^2 \ge 2xy > \frac 12 xy$.

Or...

$(x-y)^2 \ge 0$ and so $x^2 - 4xy + y^2 \ge 0$ and $x^2 + y^2 \ge 4xy \ge \frac 12 xy$ so $2x^2 + 2y^2 \ge xy$.

Or we could assume, wolog, that $x\le y$ in which case $x^2 \le xy \le y^2 < 2x^2 + 2y^2$.

Seems like way overkill in the statement.

If $x=y$ then we but $\frac x{x+2y} + \frac y{2+2x} = \frac 23$.

If we substitute $m =\frac {x+y}2$ and $d= m-x$ then we have $y = m+ d$ and we have

$\frac x{x+2y} + \frac y{y+2x} =$

$\frac {m-d}{3m+2d} + \frac {m+d}{3m - 2d}=$

$\frac {(m-d)(3m-2d) + (m+d)(3m+2d)}{(3m+2d)(3m -2d)} =$

$\frac {6m^2+4d^2}{9m^2 - 4d^2} \ge$

$\frac {6m^2}{9m^2} = \frac 23$.

That's a stronger statement:

$\frac x{x+2y} + \frac y{y+2x} \ge \frac 23$.