Inequality challenge proof day 2

I.. \(A,B,C\) are angles of a triangle. Prove that

cyccosAcycsin(A2)\sum_{\mathrm{cyc}}\cos A\leq\sum_{\mathrm{cyc}}\sin\left(\frac{A}{2}\right)

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

cyca2+b2a8+b8a3+b3+c3\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a^{8}+b^{8}}\leq a^{3}+b^{3}+c^{3}

iii. For a,b>0a,b>0, prove that (1+a)8+(1+b)8128ab(a+b)2(1+a)^{8}+(1+b)^{8}\geq 128ab(a+b)^{2}

iv. p,q,rp,q,r are distinct prime numbers such that rp3+p2+p=2rq2+q2+qrp^{3}+p^{2}+p=2rq^{2}+q^{2}+q. Find all possible values of pqrpqr.

v. If a,ba,b are integers, then prove that at least one of these expressions is an integer: a2+bb2a,b2+aa2b\frac{a^{2}+b}{b^{2}-a},\frac{b^{2}+a}{a^{2}-b}.

vi. cyclica,b,c(b+c)(a4(bc)2)ab+2bc+ac0\sum_{cyclic}^{a,b,c}\frac{(b+c)(a^{4}-(bc)^{2})}{ab+2bc+ac}\geq0.

Note by Shivam Jadhav
5 years, 10 months ago

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(1+a)8+(1+b)8(2a)8+(2b)8=256(a4+b4)(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(a4+b4)128ab(a+b)2256(a^4+b^4)\ge 128ab(a+b)^2 which is obvious by expansion.

Daniel Liu - 5 years, 10 months ago

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Q1) cosA+cosB=2cos(A+B2)cos(AB2)\cos A + \cos B = 2 \cos \left(\dfrac{A+B}{2} \right) \cos \left(\dfrac{A-B}{2} \right)

Since A,B,CA,B,C are angles of triangle. This gives us that A+B+C=π    A+B=πCA+B+C= \pi \implies A+B = \pi - C. We also have cos(AB2)1\cos \left(\dfrac{A-B}{2} \right) \leq 1

So, cosA+cosB=2cos(πC2)cos(AB2)=2sinC2cos(AB2)2sinC2\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 cosB+cosC2sinA2\cos B + \cos C \leq 2\sin \dfrac{A}{2} and cosC+cosA2sinB2\cos C + \cos A \leq 2\sin \dfrac{B}{2}.

Adding all these we get,

cyccosAcycsinA2\sum_{cyc} \cos A \leq \sum_{cyc} \sin \dfrac{A}{2}

And equality holds when A=B=C=π3A=B=C = \dfrac{\pi}{3}.

Surya Prakash - 5 years, 10 months ago

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What does question 4 ask for?

Jake Lai - 5 years, 10 months ago

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Its correct now.

Shivam Jadhav - 5 years, 10 months ago

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