# Questions tagged [buffalo-way]

Inequalities that can be proved by BW (Buffalo Way).

51 questions
1answer
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### Prove $\sum\limits_{cyc}\frac{ab}{b^{\,2}+ c^{\,2}}\geqq \frac{3}{2}$

For $a\geqq b\geqq c> 0$. Prove $$\frac{ab}{b^{\,2}+ c^{\,2}}+ \frac{bc}{c^{\,2}+ a^{\,2}}+ \frac{ca}{a^{\,2}+ b^{\,2}}\geqq \frac{3}{2}$$ I used discrim to find and I want to see a solution ...
2answers
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### Prove $2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$

For $a,\,b,\,c\geqq 0$ and $b\equiv {\rm mid}\,\{\,a,\,b,\,c\,\}$. Prove $$2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$$ Inspried from $\lceil$ Prove $k=0$ is the only non-...
1answer
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### Prove $\sum\limits_{cyc}\,\frac{a^{\,2}+ k\,b^{\,2}}{k\,a^{\,2}+ b^{\,2}}\geqq 3$

For $a\geqq b\geqq c> 0$. Prove $$\sum\limits_{cyc}\,\frac{a^{\,2}+ k\,b^{\,2}}{k\,a^{\,2}+ b^{\,2}}\geqq 3 \tag{SHED}$$ with $k= \frac{b}{c}\geqq 1$. I used SHEDtechniQ to find and I want to ...
0answers
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### Prove $\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$ with $a,\,b,\,c> 0$

Let $a,\,b,\,c$ be positive numbers. Prove that $$\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$$ I tried Holder and $\lceil$ https://...
1answer
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1answer
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### Inequality from AMM problems section

This is Problem 12024 of AMM. It asks to show that if $x,y,z$ are positive reals, and $xyz=1$, then $(x^{10}+y^{10}+z^{10})^{2}\geq 3(x^{13}+y^{13}+z^{13})$. I could show it for the particular case ...
4answers
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### If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$

If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$xw+wz+zy+yu+ux⩽\frac15.$$ I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, ...
1answer
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2answers
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### Proving $a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$ for real $a$, $b$, $c$

If $a$, $b$, $c$ are real numbers, I have to prove: $$a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$$ Since $$a^4+b^4+c^4 \ge abc(a+b+c)$$ then it is enough ...
1answer
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### Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$

If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
2answers
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1answer
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