# Questions tagged [uvw]

This is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables.

20 questions
3answers
6k views

### How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt{{\frac{abc+bcd+cda+dab}{4}}}$

How to prove this inequality $$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt{{\frac{abc+bcd+cda+dab}{4}}} ?$$ Thanks
4answers
902 views

### How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$

let $a,b,c>0$,and such $a+b+c=3$, show that $$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$ I think this inequality use this $$ab\le\dfrac{(a+b)^2}{4}$$
3answers
666 views

### Hard inequality $(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4}$

I need to prove or disprove the following inequality: $$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4}$$ For $x,y,z \in \mathbb R^+$. I found no counter ...
4answers
380 views

### Prove $\sum{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$

Let $x,y,z>0$: Prove that: $\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}$ I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
3answers
322 views

2answers
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5answers
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### If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ [closed]

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ This can be easily done by calculas but is there any way to do do this by algebra
1answer
224 views