# 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
6 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: $\sin A+\sin B+\sin C\le \dfrac{3\sqrt{3}}{2}$

$\Box$

- 6 years, 4 months ago

well explained

- 6 years, 4 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 :')

- 6 years, 4 months ago

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

- 6 years, 4 months ago

- 6 years, 4 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 :)

- 6 years, 4 months ago

Thanks really helpful great job you proved the second one

- 6 years, 4 months ago

Yes definitely tag him. :D

- 6 years, 4 months ago