Help with proof!

1) Prove that for any \triangle ABC, we have sinA+sinB+sinC332.\sin{A}+\sin{B}+\sin{C}\leq \frac{3\sqrt{3}}{2}.

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

aa2+8bc+bb2+8ca+cc2+8ab1\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1

Note by Mardokay Mosazghi
5 years ago

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Solution for first one:

Note that sinx\sin x is a concave function in the range 0xπ0\le x\le \pi. Thus, we use Jensen's Inequality where F(x)=sinxF(x)=\sin x to get sin(13(A+B+C))13sinA+13sinB+13sinC\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 33 gives 3sin(13(A+B+C))sinA+sinB+sinC3\sin\left(\dfrac{1}{3}(A+B+C)\right)\ge \sin A+\sin B+\sin C

However, A+B+C=πA+B+C=\pi. Thus, 3sin(13(A+B+C))=3sin(π3)=3323\sin\left(\dfrac{1}{3}(A+B+C)\right)=3\sin\left(\dfrac{\pi}{3}\right)=\dfrac{3\sqrt{3}}{2}

The result follows: sinA+sinB+sinC332\sin A+\sin B+\sin C\le \dfrac{3\sqrt{3}}{2}

\Box

Daniel Liu - 5 years ago

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well explained

Mardokay Mosazghi - 5 years ago

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

We want to prove cycaa2+8bc1\sum_{cyc}\dfrac{a}{\sqrt{a^2+8bc}}\ge 1

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

By Cauchy, (cycaa2+8bc)(cycaa2+8bc)(a+b+c)2=9\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 cycaa2+8bc\displaystyle\sum_{cyc}a\sqrt{a^2+8bc}, we see that we want to prove 9cycaa2+8bc1\dfrac{9}{\sum\limits_{cyc}a\sqrt{a^2+8bc}}\ge 1 or equivalently cycaa2+8bc9\sum\limits_{cyc}a\sqrt{a^2+8bc}\le 9

Squaring both sides, we have (cycaa2+8bc)281\left(\sum_{cyc}a\sqrt{a^2+8bc}\right)^2\le 81

Now use Cauchy again to obtain (cycaa2+8bc)2(a+b+c)(cyca(a2+8bc))81\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=3a+b+c=3, the inequality becomes cyca3+8abc27\sum_{cyc}a^3+8abc\le 27 after some simplifying.

But this equals (a+b+c)33(syma2b)+18abc27(a+b+c)^3-3\left(\sum_{sym}a^2b\right)+18abc\le 27 and since a+b+c=3a+b+c=3 we just want to prove (syma2b)6abc\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 - 5 years ago

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

Sudeep Salgia - 5 years ago

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@Daniel Liu

Mardokay Mosazghi - 5 years ago

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Yes definitely tag him. :D

Finn Hulse - 5 years 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 - 5 years ago

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Thanks really helpful great job you proved the second one

Mardokay Mosazghi - 5 years ago

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