An Inequality with Just Two Variable VIII

Problem

Solution 1

Let $\displaystyle f(a,b)=\frac{a^2}{b+2}+\frac{b^3}{a+2}+(2-a)b^2.\,$ Then $f(0,0)=0;\,$ $f(a\gt 0,0)=\displaystyle \frac{a^2}{2}\le 12;\,$ $f(0,b\gt 0)=\displaystyle \frac{b^3}{2}+2b^2\le 12.\,$ Suppose $a,b\gt 0:$

If $b\le 1,\,$

$f(a,b)\le\displaystyle \frac{a^2}{a}+\frac{b^2}{b}+(2-a)=b^2+2\le 12.$
If $b\gt 1,\,$

$\displaystyle \begin{align} f(a,b)-f(0,b)&=\frac{a^2}{b+2}+\frac{b^3}{a+2}+(2-a)b^2-\left(\frac{b^3}{2}+2b^2\right)\\ &=\frac{a^2}{b+2}-ab^2-\frac{ab^3}{2(a+2)}\\ &=a\left(\frac{a}{b+2}-b^2-\frac{b^3}{2(a+2)}\right)\\ &\le a\left(1-b^2-\frac{b^3}{2(a+2)}\right)\lt 0, \end{align}$

implying that $f(a,b)\lt f(0,b)=\displaystyle \frac{b^3}{2}+2b^2\le 12.\,$

Finally, $f(a,b)\le 12,\,$ with equality if and only if $a=0\,$ and $b=2.$

Solution 2

We have

$\displaystyle \begin{align} &\frac{a^2}{b+2}+\frac{b^3}{a+2}+(2-a)b^2\le 12\,\Leftrightarrow\\ &\frac{a^2}{b+2}-4+2b^2-8+\frac{a^2}{b+2}-ab^2\le 0\,\Leftrightarrow\\ &\frac{(b-2)(b^2+2b+4)-4a}{a2}+2(b-2)(b+2)+\frac{a^2}{b+2}-ab^2\le 0\,\Leftrightarrow\\ &\frac{(b-2)(b^2+2b+4)}{a+2}+2(b-2)(b+2)+a\left(\frac{a}{b+2}-b^2-\frac{4}{a+2}\right)\le 0, \end{align}$

which is true because

$\displaystyle \frac{a}{b+2}-b^2-\frac{4}{a+2}\le\frac{a}{2}-\frac{4}{2+2}-b^2\le 0,$

because $a,b\in [0,2].\,$ Equality holds if and only if $a=0\,$ and $b=2.$

Solution 3

$\displaystyle \frac{a^2}{b+2}+\frac{b^3}{a+2}+(2-a)b^2 \leq \frac{2a}{b+2}+4+2b(2-a)$

For a fixed value of $b$, the right hand side is a linear function of $a$ with slope $2/(b+2)-2b$. The slope is non-negative when $b\in[0,\sqrt{2}-1]$ and negative when $b\in(\sqrt{2}-1,2]$.

In the regime where the slope is non-negative, $RHS\,$ attains maximum value of $4/(b+2)+4$ at $a=2$ (the largest value of $a$). This function of $b$, in turn, attains a maximum value of $6$ at $b=0$.

In the regime where the slope is negative, $RHS\,$ attains maximum value of $4+4b$ at $a=0$ (the smallest value of $a$). Clearly, this function takes a maximum value of $12$ when $b=2$ (the largest value of $b).$

Thus, $LHS \leq 12$.

Acknowledgment

This problem from his book "Algebraic Phenomenon" has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Redwane El Mellass; Solution 2 is by Richdad Phuc; Solution 3 is by Amit Itagi.

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