# Questions tagged [uvw]

The uvw method is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables as well. This tag should be used for questions that could be tackled with this method, or questions about the method itself.

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### Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}$ when $a+b+c=3.$

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}.}$$ Equality holds at $a=b=c=1$ or $a=b=0;c=3.$ I ...
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### How to prove $2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$?

Question. Let $a,b,c>0: abc=a+b+c+2.$ Prove that$$2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$$I am looking ...
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### How to prove $\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\frac{1}{2}abc+\sqrt{2}$ for $ab+bc+ca=1.$

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\frac{1}{2}abc+\sqrt{2}.}$$ Since $abc\ge 0$ and equality holds at ...
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### Prove $\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} },$ if $a+b+c+abc=4.$

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove that $$\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} .}$$ Source: ...
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### For $x+y+z=3,$ prove $\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$

Let $x,y,z\ge 0: x+y+z=3.$ Prove that$$\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$$ Here is just my thought progress. I set $0\le xy+yz+zx=q\le 3; 0\le r=xyz\le 1.$ After full ...
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### Finding $\small{\max\limits_{ab+bc+ca=abc+2}\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}.}$

If $a,b,c\ge 0: ab+bc+ca=abc+2.$ Find the maximum $$P=\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}$$ By set $a=b=c=1,$ we can see that $P= 1$. The remain is ...
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