Inequality from Memo 2012

Let $a, b, c$ be positive real numbers such that $abc=1$. Prove that:

$$\sqrt{9+16a^2}+\sqrt{9+16b^2}+\sqrt{9+16c^2} \geq 3 + 4(a+b+c)$$

 Solution

Define $x=\ln a, \; y= \ln b, \; z = \ln c$ and note that $x+y+z =0$ that is implied from the initial condition. The inequality takes a new form , namely:

$$\sum \left( \sqrt{9+16e^{2x}} - 4e^x \right) \geq 3$$

Let us define the function  $f(u)=\sqrt{9+16e^{2u}} - 4e^u$ which can easily be proved that is convex for $u \geq 0$. Hence, because of HCF theorem we need to prove it just when $a=b$. That is, we need to prove that if $a^2 c =1$ then

$$2 \sqrt{9+16a^2}+\sqrt{9+16c^2} \geq 3+4(2a+c)$$

Squaring both sides it remains to prove

$$\sqrt{(9+16a^2)(9+16c^2)} \geq 16ac+12a+6c-9$$

or equivelantly (using the fact that $c=1/a^2$)

$$\frac{12}{a^4}(a-1)^2(18a^3+4a^2+2a+9) \geq 0$$

which is obviously true. Equality is valied if $a=b=c=1$.

The exercise can also be found at aops.com