Articles of uvw

Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$

For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}$$ We have: $P=\left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{\left(z+1\right)^2\left(z^2+1\right)}+\frac{1}{\left(y+1\right)^2\left(y^2+1\right)}+\frac{1}{\left(x+1\right)^2\left(x^2+1\right)}\right)$ $\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{1}{2\left(z^2+1\right)^2}+\frac{1}{2\left(x^2+1\right)^2}+\frac{1}{2\left(y^2+1\right)^2}\right)$ $\ge \left(\left(x+1\right)\left(y+1\right)\left(z+1\right)\right)^2\left(\frac{9}{2\left(\left(z^2+1\right)^2+\left(y^2+1\right)^2+\left(x^2+1\right)^2\right)}\right)$ I can’t continue. Help

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\over{ab}}$

If $a^2+b^2=1$, where $a>0$ and $b>0$, 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

If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The “building blocks” of the constraints and relations I have been looking at are: $$ \sum_{\mbox{cyc}} 1 \equiv 3 \\ \sum_{\mbox{cyc}} a \\ \sum_{\mbox{cyc}} ab \\ \sum_{\mbox{cyc}} a^2 […]

How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$

Question: let $x,y,z>0$ and such $xyz=1$, show that $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$ My idea: use AM-GM inequality $$x^3+x^3+1\ge 3x^2$$ $$y^3+y^3+1\ge 3y^2$$ $$z^3+z^3+1\ge 3z^2$$ so $$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger

Combined AM GM QM inequality

I came across this interesting inequality, and was looking for interesting proofs. $x,y,z \geq 0$ $$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$ Addendum. In general, when is $$ a\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+b\sqrt [3]{xyz}\leq (a+b)\left(\frac{x+y+z}{3}\right) $$ true?

Find the maximum value of the expression…

Find the maximum value of the expression : ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$ where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$. Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$. How can I actually solve the problem without making unnecessary assumptions ?

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 $$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} ?$$ Thanks

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 examples, so I think it should be true. I tried Cauchy-Schwarz, but I didn’t get anything useful. Is it possible to prove this inequality without using brute force methods like Bunching and Schur? This […]