Prove Inequality: $(x+y)^2/2+ (x+y)/4 \ge x\sqrt{y}+y\sqrt{x}$

[anemone]

Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.

 

[lfdahl]

Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.

Hint:

Spoiler

request for a small hint

 

[anemone]

Hint:

Spoiler

request for a small hint

Here it goes:

Spoiler

Try to prove it by breaking it down to the cases where both $x,\,y$ are less than zero, equal zero or greater than zero...and that's my solution.

But, a friend of mine proved it in the more elegant manner, he used the AM-GM method that completely simplifies the method of proving very elegantly.

 

[Albert1]

Here it goes:

Spoiler

Try to prove it by breaking it down to the cases where both $x,\,y$ are less than zero, equal zero or greater than zero...and that's my solution.

But, a friend of mine proved it in the more elegant manner, he used the AM-GM method that completely simplifies the method of proving very elegantly.

Spoiler

$x,y$ can not be negative, or $\sqrt x,\sqrt y $ will be meanless
so $x\geq 0, y\geq 0$ under this condiion $AM-GM$ method can be used

 

[anemone]

Spoiler

$x,y$ can not be negative, or $\sqrt x,\sqrt y $ will be meanless
so $x\geq 0, y\geq 0$ under this condiion $AM-GM$ method can be used

Spoiler

Ah, I was juggling too much things this morning my time that I didn't realize my hint was partly wrong. Sorry about that.

My proof is to break down the problem into three cases where both are zero, and then $x\ge y\ge 1$ and last, $0<x\le y <1$.

Yes, AM-GM could be use, but in my method, I have to also rely on the calculus method to complete the proof. But if one can see how to rewrite the problem, one could solve it very elegantly with only AM-GM inequality formula.

 

[Albert1]

Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.

use $AM-GM$ only

Spoiler

for $x\geq 0, y\geq 0$
let $A=\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}$
and $B=x\sqrt{y}+y\sqrt{x}$
$A\geq 2xy+\dfrac {x+y}{4}\geq \sqrt {2x^2y+2y^2x}$
$\therefore A^2=x^2y+y^2x+x^2y+y^2x\geq x^2y+y^2x+2xy\sqrt {xy}=B^2$
and we have $A\geq B$

 

[anemone]

use $AM-GM$ only

Spoiler

for $x\geq 0, y\geq 0$
let $A=\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}$
and $B=x\sqrt{y}+y\sqrt{x}$
$A\geq 2xy+\dfrac {x+y}{4}\geq \sqrt {2x^2y+2y^2x}$
$\therefore A^2=x^2y+y^2x+x^2y+y^2x\geq x^2y+y^2x+2xy\sqrt {xy}=B^2$
and we have $A\geq B$

Very good, Albert! And thanks for participating!