Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$

Question

Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$

My attempt,

By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$

$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$

$$=4ab+8ac+8bc+16c\sqrt{ab}$$

$$\frac{(a+b)^2+(a+b+4c)^2}{abc}\cdot (a+b+c)\geq\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}\cdot (a+b+c)$$

$$=(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{ab}})(a+b+c)$$

$$=8(\frac{1}{2c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}})(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c)$$

$$\geq 8(5\sqrt[5]{\frac{1}{2a^2b^2c}})(5\sqrt[5]{\frac{a^2b^2c}{2^4}})=100$$

Hence the largest constant $k$ is $100$

Is my answer correct? Is there another way to solve it? Thanks in advance.

Answer 1

I think you mean that $a$, $b$ and $c$ are positives.

For positive variables your solution is true, I think.

Another way:

Let $a+b=4xc$.

Hence, by AM-GM $$\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{abc}\geq\frac{((a+b)^2+(a+b+4c)^2)(a+b+c)}{\left(\frac{a+b}{2}\right)^2c}=$$ $$=\frac{(16x^2+(4x+4)^2)(4x+1)}{4x^2}=\frac{4(2x^2+2x+1)(4x+1)}{x^2}\geq$$ $$\geq\frac{4\cdot5\sqrt[5]{(x^2)^2\cdot x^2\cdot1}\cdot5\sqrt[5]{x^4\cdot1}}{x^2}=100.$$ The equality occurs for $x=1$, which says that the answer is $100.$

Done!

Answer 2

One more way...

Noting that replacing $a, b$ with $\frac{a+b}2, \frac{a+b}2$ leaves RHS unchanged but increases the LHS, we have to only check for the case $a=b$. Further as the inequality is homogeneous in $a, b, c$; WLOG we may set $a=1$. Hence we need only look for the minimum of the univariate $$f(c) = (4+(2+4c)^2)\cdot \frac{2+c}c$$ This is easily done using calculus, or by noting $f(c) = \dfrac{4(4+c)(2c-1)^2}c + 100 \geqslant 100$.