<s>[INST] Prove $4(a^2 + b^2 + c^2) + 9(abc)^2 \geq 21.$ Could anyone advise me how to prove this inequality? Hints will suffice, thank you.
Let $a,b,c \in [0, \infty)$ such that $ab+bc+ ca =3.$ Prove $4(a^2 + b^2 + c^2) + 9(abc)^2 \geq 21.$
My attempt: By AM-GM inequality, $(abc)^{2}\leq 1$ and $a^2 + b^2 + c^2 \geq 3(abc)^{\frac{2}{3}}.$
 [/INST] If we assume the following condition (without loss of generality) $a\geq b \geq c$ then we have :
$$4(a^2+b^2+c^2)+9(abc)^2\geq 4(a^2+b^2+c^2)+9(c)^6$$
With your initial condition $ab+bc+ca=3$ the RHS of the inequality  can be rewritten as follows :
$$4((\frac{3-ab}{a+b})^2+(a+b)^2-2ab)+9(\frac{3-ab}{a+b})^6$$
Now we put $a+b=cst=x$ and remark that the inequality is minimal when $ab$ is maximal where the maximum value of $ab$ in term of $x$ is $\frac{x^2}{4}$
So we have to study this function :
$$f(x)=4((\frac{3-\frac{x^2}{4}}{x})^2+(x)^2-2\frac{x^2}{4})+9(\frac{3-\frac{x^2}{4}}{x})^6$$
But $max(a+b)=2$ so we have to study on the domain $]0;2]$ the function $f(x)$
It's clear that the function is decreasing on his domain and furthermore we have :
$$f(x)\geq f(2)\geq 21$$
So to conclude we have :
$$4(a^2+b^2+c^2)+9(abc)^2\geq 4(a^2+b^2+c^2)+9(c)^6\geq 21$$
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