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

Note by Shivam Jadhav
10 months, 2 weeks 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 \(\cos C + \cos A \leq 2\sin \dfrac{B}{2}\).

Adding all these we get,

\[\sum_{cyc} \cos A \leq \sum_{cyc} \sin \dfrac{A}{2}\]

And equality holds when \(A=B=C = \dfrac{\pi}{3}\). Surya Prakash · 10 months, 2 weeks 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. Daniel Liu · 10 months, 1 week ago

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What does question 4 ask for? Jake Lai · 10 months, 2 weeks ago

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@Jake Lai Its correct now. Shivam Jadhav · 10 months, 2 weeks ago

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