<s>[INST] How prove this inequality $3b+8c+abc\le 12$ if $a^2+4b^2+9c^2=14$ let $a,b,c>0$ and such $$a^2+4b^2+9c^2=14$$show that
$$3b+8c+abc\le 12$$
My idea: since
\begin{align*}3b+8c+abc&=3b+c(8+ab)=3b+\dfrac{1}{9}\cdot 9c(8+ab)\le 3b+\dfrac{1}{9}\cdot\dfrac{1}{4}[81c^2+(ab+8)^2]\\
&=3b+\dfrac{1}{36}[126-9a^2-36b^2+a^2b^2+16ab+64]\\
&=\dfrac{1}{36}(a^2b^2-9a^2-36b^2+16ab+108b+190)
\end{align*}
then we only prove this inequality
$$a^2b^2-9a^2-36b^2+16ab+108b+190\le 36\times 12=432$$
$$\Longleftrightarrow a^2b^2-9a^2-36b^2+16ab+108b\le 242$$
then I can't prove it,Thank you for you help
 [/INST] $\dfrac{b^2+1}{2}\ge b \implies \dfrac32(b^2+1)\ge 3b$  
$\dfrac{c^2+1}{2}\ge c \implies 4(c^2+1)\ge 8c$  
Adding these two equations,  
$4c^2+\dfrac32b^2+\dfrac{11}{2}\ge 3b+4c$  
Now, $12-\dfrac{11}{2}=\dfrac{13}{2}$  
We know, $7>\dfrac{13}{2}$ and $7=\dfrac{(a^2+4b^2+9c^2)}{2}$  
So, if we can now prove that $\dfrac12a^2+\dfrac12b^2+\dfrac12c^2\ge abc+\dfrac12$, Then we are done.  
Can you do it from here?  
I think weighted AM-GM should help.  
EDIT: As contributed by Barto, here is the complete solution of the question.  
To prove, $a^2+b^2+c^2\ge 2abc+1$, we can re-write this equation as,  
$14(a^2+b^2+c^2)\ge 28abc+a^2+4b^2+9c^2 \implies 13a^2+10b^2+5c^2\ge 28abc$  
By weighted AM-GM,  $13a^2+10b^2+5c^2\ge28\sqrt[28]{a^{26}b^{20}c^{10}}\ge 28abc$ from @MarkBennett's comment.  
Hence, proved.
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