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I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated

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    $\begingroup$ I presume you want $a$, $b\ge0$? $\endgroup$ – Lord Shark the Unknown Jan 5 at 6:14
  • $\begingroup$ This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ – Carl Mummert Jan 11 at 13:55
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Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets $$\frac15\ge\sqrt[5]{\frac a2\frac a2\frac b3\frac b3\frac b3}$$ etc.

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Just using algebra.

Using $b=1-a$, you are looking for the maximum of function $$f(a)=a^2(1-a)^3$$ Compute the derivatives $$f'(a)=-a(1-a)^2 (5 a-2)$$ $$f''(a)=2(1-a)(10 a^2-8 a+1)$$ The first derivative cancels at $a=0$, $a=1$ and $a=\frac 25$. For the first two, $f(a)=0$. For the last one $f\left(\frac{2}{5}\right)=\frac{108}{3125}$ and $f''\left(\frac{2}{5}\right)=-\frac{18}{25} <0$ confirms that this is a maximum value.

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  • $\begingroup$ Although it's not the most elegant, this is by far the most straightforward answer! $\endgroup$ – Toby Mak Jan 5 at 6:55
  • $\begingroup$ You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier. $\endgroup$ – Toby Mak Jan 5 at 7:02
  • $\begingroup$ Is $f''(x)$ required at all? $\endgroup$ – Shubham Johri Jan 5 at 7:03
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    $\begingroup$ @ShubhamJohri; Just to confirm that this is a maximum. $\endgroup$ – Claude Leibovici Jan 5 at 7:08
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Does not exist. Try $b\rightarrow+\infty$.

For non-negatives $a$ and $b$ by AM-GM we obtain: $$a^2b^3=2^23^3\cdot\left(\frac{a}{2}\right)^2\left(\frac{b}{3}\right)^3\leq2^23^3\left(\frac{2\cdot\frac{a}{2}+3\cdot\frac{b}{3}}{5}\right)^5=\frac{108}{3125}.$$ The equality occurs for $\frac{a}{2}=\frac{b}{3},$ which says that we got a maximal value.

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$$f(x)=(1-x)^2\cdot x^3=x^5-2x^4+x^3$$ $$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$ $$f'(x) = 0 \implies x \in \{0, 1, \frac 35\}$$ $$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$ $$f''(0)=0, f''(1)>0, f''(\frac 35)<0$$

Hence $b=\frac 35, a=\frac 25$, if we assume $a,b > 0$

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  • $\begingroup$ I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same. $\endgroup$ – Mark Bennet Jan 5 at 6:57
  • $\begingroup$ We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$ $\endgroup$ – Shubham Johri Jan 5 at 7:06
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Let Constraint function be $$C(a,b)= a+b-1=0 $$ and the Object function that needs maximization be $$G(a,b)=a^2b^3 $$ we have with partial differentiation of combined Lagrangian: $$ C(a,b)- \lambda G(a,b) $$ condition to evaluate the multiplier $\lambda$ $$\dfrac {\dfrac{\partial C(a,b)}{\partial a} } {\dfrac{\partial C(a,b)}{\partial b} } =\dfrac {\dfrac{\partial G(a,b)}{\partial a}} {\dfrac{\partial G(a,b)}{\partial b} } $$ $$\dfrac{2a^2b^3}{3a^3b^2} =1\quad \rightarrow \dfrac{a}{b} = \dfrac{2}{3} $$

That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.

After plugging in these values Object function maximum value is:

$$ \dfrac{27a^5}{8} =\dfrac{4b^5}{9}. $$

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