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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

math110
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  • I don't know, but do you think $$(a+2b+3c)^2=a^2+4b^2+9c^2+4ab+6ac+12bc$$ might help? – Guy Mar 15 '14 at 10:36
  • I want to use AM/GM $\frac {14}{14}=\frac {a^2+4b^2+9c^2}{14}\ge\sqrt[14]{a^2b^8c^{18}}$ which gives $ab^4c^9\le 1$. And $ab^4c^9=(abc)b^3c^8$, but it doesn't quite get there directly - someone will probably spot a weighted average solution along these lines. – Mark Bennet Mar 15 '14 at 11:01

1 Answers1

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$\dfrac{b^2+1}{2}\ge b \implies \dfrac32(b^2+1)\ge 3b$

$\dfrac{c^2+1}{2}\ge c \implies 4(c^2+1)\ge 8c$

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.

Hawk
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    To prove $a^2+b^2+c^2\geq2abc+1$, you can rewrite this as $14(a^2+b^2+c^2)\geq28abc+a^2+4b^2+9c^2$, which is equivalent to $13a^2+10b^2+5c^2\geq28abc$. By AM-GM, $LHS\geq28\sqrt[28]{a^{26}b^{20}c^{10}}$ which is at least (see Mark Bennet's comment why $ab^4c^9\leq1$) $28\sqrt[28]{a^{26}b^{20}c^{10}\cdot(ab^4c^9)^2}=28abc$. Perhaps you can add this to your answer. – Bart Michels Mar 15 '14 at 11:56
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    @barto thank you a lot for the help. you helped to complete the solution. – Hawk Mar 15 '14 at 12:48