Let $a,b,c>0$, find the $\min(a^4+b^2+c)$ if $abc=1$
I found myself struggling with this problem, here is my solution:
Claim: $\min(a^4+b^2+c)=3$
Proof: We want to prove first that $a^4+b^2+c\geq a^2+b^2+c^2$: (1)
We note that $c=\frac{1}{ab}$, wlog $a \geq b \geq c$, then $\frac{1}{a^2}≤c=\frac{1}{ab}≤\frac{1}{b^2}$ $\\$ So $c \geq \frac{1}{a^2}$. We get that $a \geq \frac{1}{\sqrt{c}}$. Let $a=\frac{1}{\sqrt{c}}$ For convenience we set it as the minimum value and we we replace the expression (1): We get that is true for every $c>0$
So $$a^4+b^2+c\geq a^2+b^2+c^2 \geq a+b+c \geq 3$$
By AM-GM $$a^4+b^2+c=a^4+2\cdot\frac{b^2}{2}+4\cdot\frac{c}{4}\geq7\sqrt[7]{a^4\left(\frac{b^2}{2}\right)^2\left(\frac{c}{4}\right)^4}=\frac{7}{\sqrt[7]{1024}}.$$ The equality occurs for $a^4=\frac{b^2}{2}=\frac{c}{4},$ which says that we got a minimal value.