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

In summary, the conversation discusses an inequality being proven, which is $(x+y)^2/2+ (x+y)/4 \ge x\sqrt{y}+y\sqrt{x}$ and the variables involved are x and y. The inequality can be proven by expanding and simplifying or using mathematical induction. It is significant because it shows the relationship between the sum of two numbers and the square root of their product. However, there are limitations as it only holds true for positive values of x and y and may not be valid for all values. It is important to verify its validity before using it in mathematical problems.
  • #1
anemone
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Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.
 
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  • #2
anemone said:
Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.

Hint:

request for a small hint :eek:
 
  • #3
lfdahl said:
Hint:

request for a small hint :eek:

Here it goes:

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.
 
  • #4
anemone said:
Here it goes:

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.
$x,y$ 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
 
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  • #5
Albert said:
$x,y$ 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
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.
 
  • #6
anemone said:
Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.
use $AM-GM$ only
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 $A\geq B$
 
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  • #7
Albert said:
use $AM-GM$ only
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!
 

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

What is the inequality being proven?

The inequality being proven is $(x+y)^2/2+ (x+y)/4 \ge x\sqrt{y}+y\sqrt{x}$.

What are the variables in the inequality?

The variables in the inequality are x and y.

How is this inequality proven?

This inequality can be proven by expanding both sides of the equation and simplifying, or by using mathematical induction.

What is the significance of this inequality?

This inequality is significant because it shows the relationship between the sum of two numbers and the square root of their product. It can also be used to solve various mathematical problems and equations.

Are there any limitations to this inequality?

Yes, there are limitations to this inequality. It only holds true for positive values of x and y, and it may not hold true for all values of x and y. It is important to check the validity of the inequality before using it in any mathematical problems.

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