# Questions tagged [uvw]

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

121 questions
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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 discriminant to find & want to see a solution ...
0answers
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### Prove $\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$ for $x,\,y,\,z\in [\,j,\,k\,j\,]$

For $x,\,y,\,z\in [\,j,\,k\,j\,]$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ is true with $j= constant,\,k= constant> 8$ and $k_{\,\max}$...
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### Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$ If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ ...
4answers
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### 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 $$ab\le\dfrac{(a+b)^2}{4}$$
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### Inequalities, but working around an absolute value

So what I want to prove is $$|xy+xz+yz- 2(x+y+z) + 3| \leq |x^2+y^2+z^2-2(x+y+z)+3|$$ for $x,y,z\in \mathbb{R}$, and I'm aware that the RHS is just $|(x-1)^2+(y-1)^2+(z-1)^2|$. Now I'm able to ...
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### Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$

How one can prove the following. Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds: $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$ WLOG one can assume that $0\le a\le b\le c$. ...
4answers
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### Prove inequality $\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12$

Let $x,y,z > 0$ and $xyz=8.$ Prove that $$\frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12$$ I have tried with AM–GM inequality but no result.
2answers
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### Non - symetric inequality in 3 variables

How can we prove the following inequality for $a,b,c>0$: $$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}\leq \dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{c^2+a^2} \ ?$$
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1answer
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### Prove $(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$

Let $a,b,c,d>0$ and such $a+b+c+d=1$, show that $$(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$$ since $$a+b\ge 2\sqrt{ab}$$ I think this will not hold.because we have $256abcd\le 1$
5answers
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### Prove that $a^2+b^2+c^2+3\sqrt{a^2b^2c^2} \geq 2(ab+bc+ca).$

Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt{a^2b^2c^2} \geq 2(ab+bc+ca).$$ It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use ...
2answers
211 views

### Inequality exercise (olympiad)

For positive $a$, $b$, $c$ such that $abc=1$. Show that $$(ab+bc+ca)(a+b+c)+6\geq 5(a+b+c).$$ From the LHS, using AM-GM, we see that $(ab+bc+ca)(a+b+c)+6\geq 3(abc)^{2/3}3(abc)^{1/3}+6=15$. But ...