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If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$
trigonometry
inequality
contest-math
substitution
muirhead-inequality
How to prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq\frac{3\sqrt{3}}{4}$?
trigonometry
inequality
radicals
cauchy-schwarz-inequality
muirhead-inequality
Inequalities Of Polynomial [closed]
inequality
muirhead-inequality
schur-inequality
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