My problems #14

By $\text{AM-GM}$ inequality

$(a+b)^2+(a+b+4c)^2$

$= (a+b)^2+(a+2c+b+2c)^2 \geq \left(2\sqrt{ab^2}\right)+\left(2\sqrt{2ac} + 2\sqrt{2bc^2}\right)$

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

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

$= \left(\dfrac{4}{c}+\dfrac{8}{b}+\dfrac{8}{a}+\dfrac{16}{\sqrt{ab}}\right) \cdot$ $(a+b+c)$

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

$\left(\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b}{2}+\dfrac{b}{2}+c\right) \geq 8\left(5\sqrt[5]{\dfrac{1}{2a^2b^2c}}\right)\left(5\sqrt[5]{\dfrac{a^2b^2c}{2^4}}\right) = \boxed{100}$

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Again $\text{AM-GM}$ inequality.

Hence, largest constant $k$ is $100$

For $k=100$ equality holds if, $a=b=2c > 0$

Q.E.D