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# 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. $\sum_{cyclic}^{a,b,c}\frac{(b+c)(a^{4}-(bc)^{2})}{ab+2bc+ac}\geq0$.

2 years, 2 months ago

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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{align*} \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{align*}

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 $$A=B=C = \dfrac{\pi}{3}$$.

- 2 years, 2 months ago

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.

- 2 years, 1 month ago

What does question 4 ask for?

- 2 years, 2 months ago

Its correct now.

- 2 years, 2 months ago