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@Naitik Sanghavi
–
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

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}}$$.$

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## Comments

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TopNewestIs it 3456/3125. ?

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please write the solution

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Am i correct?

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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}}$$.$

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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.

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

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i guess u used the cauchy schwarz inequality

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