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anemone
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Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.
anemone said:Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.
lfdahl said:Hint:
request for a small hint
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.
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
use $AM-GM$ onlyanemone said:Prove $\dfrac{(x+y)^2}{2}+\dfrac{x+y}{4}\ge x\sqrt{y}+y\sqrt{x}$.
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 $A\geq B$
The inequality being proven is $(x+y)^2/2+ (x+y)/4 \ge x\sqrt{y}+y\sqrt{x}$.
The variables in the inequality are x and y.
This inequality can be proven by expanding both sides of the equation and simplifying, or by using mathematical induction.
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.
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.