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This question already has an answer here:

Let $x$, $y$ are two positive real numbers such that $x+y=2$. Then show that $x^3y^3(x^3+y^3)≤2$.

I have tried a lot using $AP$ $GP$ inequality formulas but failed. Also proceed like that, $x+y=2$ implies $(x+y)^3=8$ implies $x^3+y^3≤ 8$ since $x$ and $y$ are positive, but failed to prove. Please help me to prove this inequality.

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marked as duplicate by Martin R, Parcly Taxel, Rolf Hoyer, Leucippus, Misha Lavrov Dec 6 '17 at 0:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $t=xy$ then you can rewrite like this $$t^3(4-3t)\leq 1$$ But this is equviavlent to $$ (t-1)^2(3t^2+2t+1) = 3t^4-4t^3+1\geq 0$$

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    $\begingroup$ You can even apply AM/GM to $t,t,t,4-3t$ to get to the conclusion directly from the first line. $\endgroup$ – user491874 Dec 5 '17 at 5:47
  • $\begingroup$ This is a nice one. Thank you @John Watson $\endgroup$ – abcdmath Dec 5 '17 at 8:46
  • $\begingroup$ Actually we have more here. We see that the proposed inequality is true for all $x,y$ not necessary positive. $\endgroup$ – Aqua Dec 9 '17 at 17:06
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By AM-GM $$x^3y^3(x^3+y^3)=2x^3y^3(x^2-xy+y^2)\leq2\left(\frac{xy+xy+xy+x^2-xy+y^2}{4}\right)^4=2.$$

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  • $\begingroup$ That is essentially the same answer as @JohnWatson’s. Both are nice! $\endgroup$ – user491874 Dec 5 '17 at 6:23
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Hint:

$$x^3+y^3=x^3+(2-x)^3=8-12x+6x^2=2+6(x-1)^2$$

$$x^3y^3=\{x(2-x)\}^3=\{1-(x-1)^2\}^3$$

Now using AM-GM inequality,

$$2=\dfrac{2+6(x-1)^2+2\{1-(x-1)^2\}+2\{1-(x-1)^2\}+2\{1-(x-1)^2\}}4\ge$$

$$\sqrt[4]{\{2+6(x-1)^2\}[2\{1-(x-1)^2\}]^3}$$

$$=\sqrt[4]{8(x^3+y^3)x^3y^3}$$

$$\iff(x^3+y^3)x^3y^3\le\dfrac{2^4}8$$

The equality occurs if $2+6(x-1)^2=2\{1-(x-1)^2\}$

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Guide

1/ Set $y := 2-x$ and substitute into $f(x) := x^3y^3(x^3+y^3)$ to get a function of $x$ only. It is a polynomial of degree $8$, with leading coefficient $-6$, namely $-6x^8 + 48x^7 -152x^6 +240 x^5 -192 x^4 + 64x^3$. (Use Wolframalpha for this.)

2/ Thus, $f$ has a global maximizer at a critical point.

3/ Luckily, $df/dx$ factors into $-48x^2(x-2)^2(x-1)^3$ giving you 3 critical points: $0,1,2$. Substituting these into $f$ gives values, $0,2,0$.

4/ Therefore, $x=1$ is the global maximizer with maximum value $2$.

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Hint: if $x + y = 2$, then $y = 2 - x$ and $x^3y^3(x^3 + y^3) = x^3(2-x)^3(x^3 + (2-x)^3)$. You can then study $f(x) = x^3(2-x)^3(x^3 + (2-x)^3)$ which is infinitely differentiable (since it is a polynomial on $x$), caracterizing it's critical points. Then, if the inequality holds, you should be able to show that it's maximum is less than $2$ simply by evaluating there, previously identifying it by checking the zeores of $f'$.

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With $y = 2 - x$, your inequality becomes

$$ 6\,{x}^{8}-48\,{x}^{7}+152\,{x}^{6}-240\,{x}^{5}+192\,{x}^{4}-64\,{x}^ {3}+2 \ge 0$$

The left side factors as $f(x) = 2 (3 x^4 - 12 x^3 + 10 x^2 + 4 x + 1)(x-1)^4$.

Now $g(x) = 3 x^4 - 12 x^3 + 10 x^2 + 4 x + 1$ has derivative $g'(x) = 12 x^3 - 36 x^2 + 20 x + 4 = 4 (x-1)(3 x^2 - 6 x - 1)$ which is $0$ at $x = 1$ and $1 \pm 2 \sqrt{3}/3$. The values of $g$ there are $6, 2/3, 2/3$. Thus the minimum of $g$ is $2/3$, and so $f(x) \ge 0$ for all real $x$.

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your inequality is equivalent to $$2\, \left( 3\,{x}^{4}-12\,{x}^{3}+10\,{x}^{2}+4\,x+1 \right) \left( x -1 \right) ^{4} \geq 0$$ and $$3x^4-12x^3+10x^2+4x+1$$ Can be written as $$x^2\left(3(x-2)^2+6\right)+4x+1>0$$ since $x>0$

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