# Inequality challenge proof day 2

I.. $$A,B,C$$ are angles of a triangle. Prove that

$\sum_{\mathrm{cyc}}\cos A\leq\sum_{\mathrm{cyc}}\sin\left(\frac{A}{2}\right)$

ii. Consider $a,b,c>0, abc=1$. Prove that

$\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a^{8}+b^{8}}\leq a^{3}+b^{3}+c^{3}$

iii. For $a,b>0$, prove that $(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}$

iv. $p,q,r$ are distinct prime numbers such that $rp^{3}+p^{2}+p=2rq^{2}+q^{2}+q$. Find all possible values of $pqr$.

v. If $a,b$ are integers, then prove that at least one of these expressions is an integer: $\frac{a^{2}+b}{b^{2}-a},\frac{b^{2}+a}{a^{2}-b}$.

vi. $\sum_{cyclic}^{a,b,c}\frac{(b+c)(a^{4}-(bc)^{2})}{ab+2bc+ac}\geq0$. 5 years, 10 months ago

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q3)

$(1+a)^8+(1+b)^8\ge (2\sqrt{a})^8+(2\sqrt{b})^8=256(a^4+b^4)$ It remains to prove $256(a^4+b^4)\ge 128ab(a+b)^2$ which is obvious by expansion.

- 5 years, 10 months ago

Q1) $\cos A + \cos B = 2 \cos \left(\dfrac{A+B}{2} \right) \cos \left(\dfrac{A-B}{2} \right)$

Since $A,B,C$ are angles of triangle. This gives us that $A+B+C= \pi \implies A+B = \pi - C$. We also have $\cos \left(\dfrac{A-B}{2} \right) \leq 1$

So, \begin{aligned} \cos A + \cos B &= 2 \cos \left(\dfrac{\pi-C}{2} \right) \cos \left(\dfrac{A-B}{2} \right) \\ &= 2 \sin \dfrac{C}{2} \cos \left(\dfrac{A-B}{2} \right) \\ & \leq 2\sin \dfrac{C}{2} \end{aligned}

Similarly we get $\cos B + \cos C \leq 2\sin \dfrac{A}{2}$ and $\cos C + \cos A \leq 2\sin \dfrac{B}{2}$.

$\sum_{cyc} \cos A \leq \sum_{cyc} \sin \dfrac{A}{2}$

And equality holds when $A=B=C = \dfrac{\pi}{3}$.

- 5 years, 10 months ago

What does question 4 ask for?

- 5 years, 10 months ago

Its correct now.

- 5 years, 10 months ago