Prove that $a^2(b^2+4)+b^2(c^2+4)+c^2(a^2+4) \geq 15$

Question

Let $$a, b, c$$ be real numbers with the property that $$ a+b+c=3 $$.

Prove that $a^2(b^2+4)+b^2(c^2+4)+c^2(a^2+4) \geq 15$

Initially, I thought to use Cauchy-Schwarz Inequality and simplify.

$(a^2+b^2+c^2)^2 \geq (ab+bc+ac)^2$ which after some calulation and simplifications gives us: $a^4+b^2+c^4 \geq 2abc(a+b+c)$.

Since from the hypothesis, the sum is 3, we get: $a^4+b^2+c^4 \geq 6abc$.

However, here I get stuck. Any idea would be really apreciated!

Answer 1

Proof.

We can rewrite the inequality as $$a^2(b^2+4)+b^2(c^2+4)+c^2(a^2+4)\ge (a+b+c)^2+6$$ or

$$a^2b^2-2ab+1+b^2c^2-2bc+1+c^2a^2-2ca+1+3\left[(a^2-2a+1)+(b^2-2b+1)+(c^2-2c+1)\right]\ge 0$$ It's $$(ab-1)^2+(bc-1)^2+(ca-1)^2+3\left[(a-1)^2+(b-1)^2+(c-1)^2\right]\ge 0.$$ Hence, the proof is done. Equality holds iff $a=b=c=1.$