Question

Let x, y be positive reals such that x + y = 2. Prove that :-

$\sf{{x}^{3}{y}^{3}({x}^{3} + {y}^{3}) \leq 2}$

Answer 1

By $\sf{AM\geq GM}$ inequality, we have,

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{x^3+y^3+(x^3+y^3)}{3}}$

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{2(x^3+y^3)}{3}\quad\quad\dots(1)}$

Given that,

$\sf{\longrightarrow x+y=2}$

$\sf{\longrightarrow y=2-x}$

Let us define a real valued function $\sf{f}$ such that,

$\sf{\longrightarrow f(x)=x^3+(2-x)^3}$

$\sf{\longrightarrow f(x)=x^3+8-12x+6x^2-x^3}$

$\sf{\longrightarrow f(x)=2(3x^2-6x+4)}$

Consider the quadratic equation $\sf{3x^2-6x+4.}$

Here the coefficient of $\sf{x^2,\ 3>0.}$

Taking the discriminant,

$\sf{\longrightarrow D=(-6)^2-(4\times3\times4)}$

$\sf{\longrightarrow D=36-48}$

$\sf{\longrightarrow D=-12<0}$

Now $\sf{a>0}$ and $\sf{D<0.}$ Thus $\sf{3x^2-6x+4>0}$ and so will $\sf{f(x)}$ be. So $\sf{f(x)}$ should have a minimum.

To find minimum we've to equate derivative of $\sf{f}$ wrt $\sf{x}$ to zero, i.e.,

$\sf{\longrightarrow f'(x)=0}$

$\sf{\longrightarrow 2(6x-6)=0}$

$\sf{\longrightarrow x=1}$

So $\sf{f(x)}$ is minimum at $\sf{x=1.}$

At $\sf{x=1,}$

$\sf{\longrightarrow f(1)=1^3+(2-1)^3}$

$\sf{\longrightarrow f(1)=2}$

Therefore,

$\sf{\longrightarrow f(x)\geq2}$

$\sf{\longrightarrow x^3+(2-x)^3\geq2}$

Since $\sf{y=2-x,}$

$\sf{\longrightarrow x^3+y^3\geq2}$

Thus (1) becomes,

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{2\times2}{3}}$

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{4}{3}}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq\left(\dfrac{4}{3}\right)^3}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq\dfrac{64}{27}}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq2.37}$

Or,

$\sf{\longrightarrow\underline{\underline{x^3y^3(x^3+y^3)\leq2}}}$

Hence Proved!

Answer 2

By $\sf{AM\geq GM}$ inequality, we have,

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{x^3+y^3+(x^3+y^3)}{3}}$

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{2(x^3+y^3)}{3}\quad\quad\dots(1)}$

Given that,

$\sf{\longrightarrow x+y=2}$

$\sf{\longrightarrow y=2-x}$

Let us define a real valued function $\sf{f}$ such that,

$\sf{\longrightarrow f(x)=x^3+(2-x)^3}$

$\sf{\longrightarrow f(x)=x^3+8-12x+6x^2-x^3}$

$\sf{\longrightarrow f(x)=2(3x^2-6x+4)}$

Consider the quadratic equation $\sf{3x^2-6x+4.}$

Here the coefficient of $\sf{x^2,\ 3>0.}$

Taking the discriminant,

$\sf{\longrightarrow D=(-6)^2-(4\times3\times4)}$

$\sf{\longrightarrow D=36-48}$

$\sf{\longrightarrow D=-12<0}$

Now $\sf{a>0}$ and $\sf{D<0.}$ Thus $\sf{3x^2-6x+4>0}$ and so will $\sf{f(x)}$ be. So $\sf{f(x)}$ should have a minimum.

To find minimum we've to equate derivative of $\sf{f}$ wrt $\sf{x}$ to zero, i.e.,

$\sf{\longrightarrow f'(x)=0}$

$\sf{\longrightarrow 2(6x-6)=0}$

$\sf{\longrightarrow x=1}$

So $\sf{f(x)}$ is minimum at $\sf{x=1.}$

At $\sf{x=1,}$

$\sf{\longrightarrow f(1)=1^3+(2-1)^3}$

$\sf{\longrightarrow f(1)=2}$

Therefore,

$\sf{\longrightarrow f(x)\geq2}$

$\sf{\longrightarrow x^3+(2-x)^3\geq2}$

Since $\sf{y=2-x,}$

$\sf{\longrightarrow x^3+y^3\geq2}$

Thus (1) becomes,

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{2\times2}{3}}$

$\longrightarrow\sqrt[\sf{3}]{\sf{x^3y^3(x^3+y^3)}}\leq\sf{\dfrac{4}{3}}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq\left(\dfrac{4}{3}\right)^3}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq\dfrac{64}{27}}$

$\sf{\longrightarrow x^3y^3(x^3+y^3)\leq2.37}$

Or,

$\sf{\longrightarrow\underline{\underline{x^3y^3(x^3+y^3)\leq2}}}$

Hence Proved!

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