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
4 years, 4 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: \[\sin A+\sin B+\sin C\le \dfrac{3\sqrt{3}}{2}\]

\(\Box\)

Daniel Liu - 4 years, 4 months ago

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well explained

Mardokay Mosazghi - 4 years, 4 months ago

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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 :')

Daniel Liu - 4 years, 4 months ago

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Comment deleted May 21, 2014

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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?

Daniel Liu - 4 years, 4 months ago

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@Daniel Liu I don't understand your comment. Where did the power 5/6 come from?

Calvin Lin Staff - 4 years, 4 months ago

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@Calvin Lin 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)

Daniel Liu - 4 years, 4 months ago

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@Daniel Liu Great job. That is impressive.

I recognized it as IMO 2001. Look up the holders approach, it is pretty spectacular and insightful.

Calvin Lin Staff - 4 years, 4 months ago

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Hint for the first one. Use Jensen's inequality on \(\sin x\).

Sudeep Salgia - 4 years, 4 months ago

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@Daniel Liu

Mardokay Mosazghi - 4 years, 4 months ago

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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 :)

Daniel Liu - 4 years, 4 months ago

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Thanks really helpful great job you proved the second one

Mardokay Mosazghi - 4 years, 4 months ago

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Yes definitely tag him. :D

Finn Hulse - 4 years, 4 months ago

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