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# Help

What is the maximum value of $$x^2 y^3$$ for positive numbers $$x$$ and $$y$$ satisfying $$x+y=2$$?

Note by Abhishek Alva
1 year, 3 months ago

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Is it 3456/3125. ?

- 1 year, 3 months ago

- 1 year, 3 months ago

Am i correct?

- 1 year, 3 months ago

yes u are write the solution

- 1 year, 3 months ago

Apply Am gm on x,x,2y/3,2y/3,2y/3.

- 1 year, 3 months ago

a/(a^3+b^2+c )+b/(b^3+c^2+a)+c/(c^3+a^2+b) if a+b+c=3 determine the largest value of the equation given that a b c are positive real numbers

- 1 year, 3 months ago

i not getting this problem

- 1 year, 3 months ago

I got the answer to be $$\dfrac{676\sqrt{26}}{3125}$$ which is very close to your answer.First we get the max. value of $$xy$$ to be $$1$$ and $$x^2+y^2$$ to be $$2$$ by $$AM-GM$$ inequality.Now we must find the max value of $$2x+3y.$$We have an inequality which states that $$ax+by\leq \sqrt{a^2+b^2}\times\sqrt{x^2+y^2}.$$Well I dont know the proof for this inequality.Hence, $$2x+3y\leq\sqrt{2^2+3^2}\times\sqrt 2=\sqrt{26}.$$ Next we apply AM-GM inequality for $$2x+3y.$$We split the terms and apply the inequality$$.\dfrac{x+x+y+y+y}{5}\geq\sqrt[5]{x^2y^3}\Rightarrow \dfrac{2x+3y}{5}\geq\sqrt[5]{x^2y^3}.$$Therefore,$$\dfrac{\sqrt{26}}{5}\geq\sqrt[5]{x^2y^3}.$$Now raise both sides by power of $$5.$$We get the maximum value of $$x^2y^3$$ to be $$\boxed{\dfrac{676\sqrt{26}}{3125}}$$$$.$$

- 1 year, 3 months ago

i guess u used the cauchy schwarz inequality

- 1 year, 3 months ago

When you applied am gm on x,x,y,y,y the equality holds when x=y which implies that max value of x²y³ is 1.I think this was ur mistake @Ayush Rai.

- 1 year, 3 months ago

But it is very close to the answer.Can u point the error?

- 1 year, 3 months ago

I already did iny other comment

- 1 year, 3 months ago