How Find this maximum $P=\frac{4}{\sqrt{a^2+b^2+c^2+4}}-\frac{9}{(a+b)\sqrt{(a+2c)(b+2c)}}$

Question

let $a,b,c>0$, find the maximum $$P=\dfrac{4}{\sqrt{a^2+b^2+c^2+4}}-\dfrac{9}{(a+b)\sqrt{(a+2c)(b+2c)}}$$

I think this inequality we can use AM-GM inequality to solve it,and Now first we must sure this equality when $a,b,c$ hold maximum

Thank you

Answer 1

First, we have by AM-GM: $$(a+b)\sqrt{(a+2c)(b+2c)} \le (a+b)\frac{a+b+4c}2 = \frac{a^2}2 + \frac{b^2}2+ab+2ac+2bc \le 2(a^2+b^2+c^2)$$

where there is equality iff $a=b=c$. So let $s = a^2+b^2+c^2$. Then we have: $$P \le \frac4{\sqrt{s+4}}- \frac9{2s}$$

Using one variable calculus, we can find the derivative condition for maximising the RHS, which gives $16s^4=81(s+4)^3$. This is equivalent to $(s-12)(16s^3+111s^2+360s+432)=0$, where the cubic factor cannot have any positive roots by the rule of signs.

Hence $P$ has a maximum for positive $s$ only when $s=12$. Thus $a=b=c=2$ at the maximum, and $P\le \frac58$.