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# Help with proof!

1) Prove that for any \triangle ABC, we have $$\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}.$$

2) Let a,b,c be positive real numbers. Prove that

$$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$$

Note by Mardokay Mosazghi
2 years, 10 months ago

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Solution for first one:

Note that $$\sin x$$ is a concave function in the range $$0\le x\le \pi$$. Thus, we use Jensen's Inequality where $$F(x)=\sin x$$ to get $\sin\left(\dfrac{1}{3}(A+B+C)\right)\ge \dfrac{1}{3}\sin A+\dfrac{1}{3}\sin B+\dfrac{1}{3}\sin C$

Multiplying both sides by $$3$$ gives $3\sin\left(\dfrac{1}{3}(A+B+C)\right)\ge \sin A+\sin B+\sin C$

However, $$A+B+C=\pi$$. Thus, $3\sin\left(\dfrac{1}{3}(A+B+C)\right)=3\sin\left(\dfrac{\pi}{3}\right)=\dfrac{3\sqrt{3}}{2}$

The result follows: $\sin A+\sin B+\sin C\le \dfrac{3\sqrt{3}}{2}$

$$\Box$$ · 2 years, 10 months ago

well explained · 2 years, 10 months ago

After an hour of working on the second one, I have finally proved it.

We want to prove $\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1$

Note that since this inequality is homogenous, assume $$a+b+c=3$$.

By Cauchy, $$\left(\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\right)\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)\ge (a+b+c)^2=9$$

Dividing both sides by $$\displaystyle\sum_{cyc}a\sqrt{a^2+8bc}$$, we see that we want to prove $\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1$ or equivalently $\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9$

Squaring both sides, we have $\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81$

Now use Cauchy again to obtain $\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le (a+b+c)\left(\sum_{cyc}a(a^2+8bc)\right)\le 81$

Since $$a+b+c=3$$, the inequality becomes $\sum_{cyc}a^3+8abc\le 27$ after some simplifying.

But this equals $(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27$ and since $$a+b+c=3$$ we just want to prove $\left(\sum_{sym}a^2b\right)\ge 6abc$ after some simplifying.

But that is true by AM-GM.

Thus, proved. QED. $$\Box$$

I have never proven something as complicated as this before, I feel so proud :') · 2 years, 10 months ago

Comment deleted May 21, 2014

Thanks, although I figured that out after thinking about it. I can't believe I messed up calculation at very end and though AM-GM on $$\displaystyle\sum_{sym}a^2b$$ gives $\displaystyle\sum_{sym}a^2b\ge 6(abc)^{5/6}$

But finally, I have proved it :'D

EDIT: Wow this is an IMO problem? · 2 years, 10 months ago

I don't understand your comment. Where did the power 5/6 come from? Staff · 2 years, 10 months ago

I thought that AM-GM on $$\displaystyle\sum_{sym}a^2b$$ yielded $\displaystyle\sum_{sym}a^2b\ge 6(abc)^{5/6}$but that is because I accidentally only counted $$5$$ $$a$$'s (and $$b$$'s and $$c$$'s) · 2 years, 10 months ago

Great job. That is impressive.

I recognized it as IMO 2001. Look up the holders approach, it is pretty spectacular and insightful. Staff · 2 years, 10 months ago

Hint for the first one. Use Jensen's inequality on $$\sin x$$. · 2 years, 10 months ago

@Daniel Liu · 2 years, 10 months ago

I'm actually really bad at inequalities right now; I only know AM-GM and Cauchy.

EDIT: OMG I can't believe I solved the second one! I feel very accomplished right now :) · 2 years, 10 months ago

Thanks really helpful great job you proved the second one · 2 years, 10 months ago