Proof: Let $E=\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}$. $\Rightarrow 4E=4\left( \frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)} \right)\ge $ $\frac{4a-2b-2c}{a+2\left( b+c \right)}+\frac{4b-2c-2a}{b+2\left( c+a \right)}+\frac{4c-2a-2b}{c+2\left( a+b \right)}=$ $\frac{5a}{a+2\left( b+c \right)}+\frac{5b}{b+2\left( c+a \right)}+\frac{5c}{c+2\left( a+b \right)}-3=$ $\frac{5{{a}^{2}}}{{{a}^{2}}+2\left( ba+ca \right)}+\frac{5{{b}^{2}}}{{{b}^{2}}+2\left( cb+ab \right)}+\frac{5{{c}^{2}}}{{{c}^{2}}+2\left( ac+bc \right)}-3\ge $ $\frac{5{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+4ab+4bc+4ca}-3=\frac{2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+4ab+4bc+4ca}\ge 0\Rightarrow E\ge 0$

Refinement Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge \frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca}{10\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$

pcoRO wrote: Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge 0.$ The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: \[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\]

pcoRO wrote: Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge 0.$ Generalization 1 Let $a,b,c\in \left( 0,\infty \right),\lambda \ge1$.Prove the inequality $\frac{a-\sqrt{bc}}{a+\lambda \left( b+c \right)}+\frac{b-\sqrt{ca}}{b+\lambda \left( c+a \right)}+\frac{c-\sqrt{ab}}{c+\lambda \left( a+b \right)}\ge \frac{\left( \lambda -1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)}{\lambda \left( 2\lambda +1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$

Generalization 2(april 8) Let $a,b,c,n,m,\lambda \in \left( 0,\infty \right),\lambda n+\lambda m\ge 2$.Prove the inequality $\frac{a-{{b}^{\frac{n}{n+m}}}{{c}^{\frac{m}{n+m}}}}{a+\lambda nb+\lambda mc}+\frac{b-{{c}^{\frac{n}{n+m}}}{{a}^{\frac{m}{n+m}}}}{b+\lambda nc+\lambda ma}+\frac{c-{{a}^{\frac{n}{n+m}}}{{b}^{\frac{m}{n+m}}}}{c+\lambda na+\lambda mb}\ge \frac{\left( \lambda n+\lambda m-2 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)}{\lambda \left( n+m \right)\left( \lambda n+\lambda m+1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$

ionbursuc wrote: Generalization 2(april 8) Let $a,b,c,n,m,\lambda \in \left( 0,\infty \right),\lambda n+\lambda m\ge 2$.Prove the inequality $\frac{a-{{b}^{\frac{n}{n+m}}}{{c}^{\frac{m}{n+m}}}}{a+\lambda nb+\lambda mc}+\frac{b-{{c}^{\frac{n}{n+m}}}{{a}^{\frac{m}{n+m}}}}{b+\lambda nc+\lambda ma}+\frac{c-{{a}^{\frac{n}{n+m}}}{{b}^{\frac{m}{n+m}}}}{c+\lambda na+\lambda mb}\ge \frac{\left( \lambda n+\lambda m-2 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)}{\lambda \left( n+m \right)\left( \lambda n+\lambda m+1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$ Please give solution

arqady wrote: pcoRO wrote: Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge 0.$ The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: \[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\] Please give solution

We need to prove that $\sum_{cyc}\frac{a^2-bc}{2a^2+b^2+c^2}\geq0$ for positives $a$, $b$ and $c$. We see that $\sum_{cyc}\frac{a^2-bc}{2a^2+b^2+c^2}=\frac{1}{2}\sum_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{2a^2+b^2+c^2}=$ $=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a+c}{2a^2+b^2+c^2}-\frac{b+c}{2b^2+a^2+c^2}\right)=$ $=\sum_{cyc}\frac{(a-b)^2(a^2+b^2+c^2-ab-ac-bc)}{2(2a^2+b^2+c^2)(2b^2+a^2+c^2)}\geq0$.

The following inequality is also true. Let $a,b,c$ be positive real numbers . Prove that\[\frac{a-\sqrt{bc}}{a+\sqrt{2\left( b^2+c^2 \right)}}+\frac{b-\sqrt{ca}} {b+\sqrt{2\left( c^2+a^2 \right)}}+\frac{c-\sqrt{ab}} {c+\sqrt{2\left( a^2+b^2 \right)}}\ge 0.\]

Very interesting. Let $a,b,c$ be positive real numbers . Prove that\[\frac{a}{a+2\left( b+c \right)}+\frac{b}{b+2\left( c+a \right)}+\frac{c}{c+2\left( a+b \right)}\ge \frac{3}{5}\]\[\ge \frac{\sqrt{bc}}{a+2\left( b+c \right)}+\frac{\sqrt{ca}}{b+2\left( c+a \right)}+\frac{\sqrt{ab}}{c+2\left( a+b \right)}.\]

Denote $\displaystyle A = \sum \dfrac {a}{a+2(b+c)}$ and $\displaystyle B = \sum \dfrac {\sqrt{bc}}{a+2(b+c)}$. Then, by the (Titu version of) Cauchy-Schwarz inequality \[A = \sum \dfrac {a}{a+2(b+c)} = \sum \dfrac {a^2}{a^2+2(ab+ca)}\geq \dfrac {\big ( \sum a \big )^2} {\sum a^2 + 4\sum bc} \geq \dfrac {3}{5},\] since it comes to $\displaystyle 2\sum a^2 \geq 2\sum bc$, i.e. $\displaystyle \sum (b-c)^2 \geq 0$. But by the AM-GM inequality \[B = \sum \dfrac {\sqrt{bc}}{a+2(b+c)} \leq \sum \dfrac {(b+c)/2}{a+2(b+c)} = \dfrac {1} {4}(3 - A),\] so $B \leq \dfrac {1} {4}\left (3 - \dfrac {3}{5}\right ) = \dfrac {3}{5}$, therefore $A \geq \dfrac {3}{5} \geq B$, whence $\boxed{A-B \geq 0}$. Equality clearly holds if and only if $a=b=c$.

sqing wrote: Very interesting. Let $a,b,c$ be positive real numbers .Prove that\[\frac{a}{a+2\left( b+c \right)}+\frac{b}{b+2\left( c+a \right)}+\frac{c}{c+2\left( a+b \right)}\ge \frac{3}{5}\]\[\ge \frac{\sqrt{bc}}{a+2\left( b+c \right)}+\frac{\sqrt{ca}}{b+2\left( c+a \right)}+\frac{\sqrt{ab}}{c+2\left( a+b \right)}.\] Generalization Let $a,b,c$ be positive real numbers and $x\ge 1.$ Prove that\[\frac{a}{a+x\left( b+c \right)}+\frac{b}{b+x\left( c+a \right)}+\frac{c}{c+x\left( a+b \right)}\ge \frac{3}{1+2x}\]\[\ge \frac{\sqrt{bc}}{a+x\left( b+c \right)}+\frac{\sqrt{ca}}{b+x\left( c+a \right)}+\frac{\sqrt{ab}}{c+x\left( a+b \right)}.\]

ionbursuc wrote: sqing wrote: Very interesting. Let $a,b,c$ be positive real numbers .Prove that\[\frac{a}{a+2\left( b+c \right)}+\frac{b}{b+2\left( c+a \right)}+\frac{c}{c+2\left( a+b \right)}\ge \frac{3}{5}\]\[\ge \frac{\sqrt{bc}}{a+2\left( b+c \right)}+\frac{\sqrt{ca}}{b+2\left( c+a \right)}+\frac{\sqrt{ab}}{c+2\left( a+b \right)}.\] Generalization Let $a,b,c$ be positive real numbers and $x\ge 1.$ Prove that\[\frac{a}{a+x\left( b+c \right)}+\frac{b}{b+x\left( c+a \right)}+\frac{c}{c+x\left( a+b \right)}\ge \frac{3}{1+2x}\]\[\ge \frac{\sqrt{bc}}{a+x\left( b+c \right)}+\frac{\sqrt{ca}}{b+x\left( c+a \right)}+\frac{\sqrt{ab}}{c+x\left( a+b \right)}.\] Generalization Let $a,b,c$ be positive real numbers and $x\ge 1.$ Prove that\[\frac{a-\sqrt{bc}}{a+x\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+x\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+x\left( a+b \right)}\ge 0.\]

ionbursuc wrote: pcoRO wrote: Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge 0.$ Generalization 1 Let $a,b,c\in \left( 0,\infty \right),\lambda \ge1$.Prove the inequality $\frac{a-\sqrt{bc}}{a+\lambda \left( b+c \right)}+\frac{b-\sqrt{ca}}{b+\lambda \left( c+a \right)}+\frac{c-\sqrt{ab}}{c+\lambda \left( a+b \right)}\ge \frac{\left( \lambda -1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)}{\lambda \left( 2\lambda +1 \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}$ $\Rightarrow $ Let $a,b,c\in \left( 0,\infty \right),\lambda \ge1$.Prove the inequality $\frac{a-\sqrt{bc}}{a+\lambda \left( b+c \right)}+\frac{b-\sqrt{ca}}{b+\lambda \left( c+a \right)}+\frac{c-\sqrt{ab}}{c+\lambda \left( a+b \right)}\ge 0$

Generalization 3 Let $a,b,c$ be positive real numbers and $x\ge \frac{1}{2}.$ Prove that\[\frac{a-\sqrt{bc}}{a+x\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+x\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+x\left( a+b \right)}\ge 0.\]

Let $a,b,c,d\in \left( 0,\infty \right),\lambda \ge 1$.Prove that $\frac{a-\sqrt[3]{bcd}}{a+\lambda \left( b+c+d \right)}+\frac{b-\sqrt[3]{cda}}{b+\lambda \left( c+d+a \right)}+\frac{c-\sqrt[3]{dab}}{c+\lambda \left( d+a+b \right)}+\frac{d-\sqrt[3]{abc}}{d+\lambda \left( a+b+c \right)}\ge 0$

Generalization 4 Let ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}\in \left( 0,\infty \right),\lambda \ge 1,n\ge 3$.Prove that $\frac{{{a}_{1}}-\sqrt[n-1]{\frac{P}{{{a}_{1}}}}}{{{a}_{1}}+\lambda \left( S-{{a}_{1}} \right)}+\frac{{{a}_{2}}-\sqrt[n-1]{\frac{P}{{{a}_{2}}}}}{{{a}_{2}}+\lambda \left( S-{{a}_{2}} \right)}+...+\frac{{{a}_{n}}-\sqrt[n-1]{\frac{P}{{{a}_{n}}}}}{{{a}_{n}}+\lambda \left( S-{{a}_{n}} \right)}\ge 0$,where $P={{a}_{1}}{{a}_{2}}...{{a}_{n}},S={{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}$

Generalization 5 a)Let $a,b,c,d\in \left( 0,\infty \right),\lambda \ge 1$.Prove that $\frac{a}{a+\lambda \left( b+c+d \right)}+\frac{b}{b+\lambda \left( c+d+a \right)}+\frac{c}{c+\lambda \left( d+a+b \right)}+\frac{d}{d+\lambda \left( a+b+c \right)}\ge \frac{4}{1+3\lambda }\ge $ $\frac{\sqrt[3]{bcd}}{a+\lambda \left( b+c+d \right)}+\frac{\sqrt[3]{cda}}{b+\lambda \left( c+d+a \right)}+\frac{\sqrt[3]{dab}}{c+\lambda \left( d+a+b \right)}+\frac{\sqrt[3]{abc}}{d+\lambda \left( a+b+c \right)}$ b)Let ${{a}_{1}},{{a}_{2}},...,{{a}_{n}}\in \left( 0,\infty \right),\lambda \ge 1,n\ge 3$.Prove that $\frac{{{a}_{1}}}{{{a}_{1}}+\lambda \left( S-{{a}_{1}} \right)}+\frac{{{a}_{2}}}{{{a}_{2}}+\lambda \left( S-{{a}_{2}} \right)}+...+\frac{{{a}_{n}}}{{{a}_{n}}+\lambda \left( S-{{a}_{n}} \right)}\ge \frac{n}{1+\left( n-1 \right)\lambda }\ge $ $\frac{\sqrt[n-1]{\frac{P}{{{a}_{1}}}}}{{{a}_{1}}+\lambda \left( S-{{a}_{1}} \right)}+\frac{\sqrt[n-1]{\frac{P}{{{a}_{2}}}}}{{{a}_{2}}+\lambda \left( S-{{a}_{2}} \right)}+...+\frac{\sqrt[n-1]{\frac{P}{{{a}_{n}}}}}{{{a}_{n}}+\lambda \left( S-{{a}_{n}} \right)}$, where $P={{a}_{1}}{{a}_{2}}...{{a}_{n}},S={{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}$.

arqady wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: \[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\] $\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0$ $\iff\frac{2\sqrt{bc}-2a}{2a+b+c }+\frac{2\sqrt{ca}-2b}{2b+ c+a}+\frac{2\sqrt{ab}-2c}{2c+a+b }\le 0$ $\iff\frac{2\sqrt{bc}+b+c}{2a+b+c }+\frac{2\sqrt{ca}+c+a}{2b+ c+a}+\frac{2\sqrt{ab}+a+b}{2c+a+b }\le 3$ $\iff\frac{(\sqrt{b}+\sqrt{c})^2}{2a+b+c }+\frac{(\sqrt{c}+\sqrt{a})^2}{2b+ c+a}+\frac{(\sqrt{a}+\sqrt{b})^2}{2c+a+b }\le 3 ,$ by C-S, $\frac{(\sqrt{b}+\sqrt{c})^2}{2a+b+c }+\frac{(\sqrt{c}+\sqrt{a})^2}{2b+ c+a}+\frac{(\sqrt{a}+\sqrt{b})^2}{2c+a+b }$ $\le \frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}+\frac{a}{b+a}+\frac{a}{c+a}+\frac{b}{c+b}=3 .$

Nice proof, sqing!

arqady wrote: Nice proof, sqing! This proof belongs Yanzongzhang. Yanzongzhang is a young math teacher of China's Zhejiang Province.

Very interesting. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&p=3146750#p3146750： nikoma ,19 Jul 2013, 08:45 • # 122：Here's problem by Pham Kim Hung: Let $a,b,c > 0$, prove that \[\sum_{cyc} \frac{a^2 - bc}{2a^2 + b^2 + c^2} \geq 0\] ssilwa,19 Jul 2013, 09:15 • # 123：(I see nikoma has been reading a certain book ) This is equivalent to:\[3 \ge \sum \frac{(a+b)^2}{a^2+b^2+2c^2} \] By Titu's lemma,\[ \frac{a^2}{a^2+c^2}+ \frac{b^2}{b^2+c^2} \ge \frac{(a+b)^2}{a^2+b^2+2c^2} \] Now just taking the cyclic sums gives us the desired inequality. Equality if $a=b=c$ or two equal each other and the third is zero.

See also here:Pham Kim Hung - Secrets in Inequalities (volume 1) (2007)P35

The following inequality is true. Let $a,b,c\in \left( 0,\infty \right)$.Prove that $\frac{a-\sqrt{bc}}{2a+b+c}+\frac{b-\sqrt{ca}}{2b+c+a}+\frac{c-\sqrt{ab}}{2c+a+b}\le \frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}+{{\left( \sqrt{b}-\sqrt{c} \right)}^{2}}+{{\left( \sqrt{c}-\sqrt{a} \right)}^{2}}}{2\left( a+b+c \right)}.$

arqady wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that:\[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\] sqing wrote: Very interesting. Let $a,b,c$ be positive real numbers . Prove that\[\frac{a}{a+2\left( b+c \right)}+\frac{b}{b+2\left( c+a \right)}+\frac{c}{c+2\left( a+b \right)}\ge \frac{3}{5}\]\[\ge \frac{\sqrt{bc}}{a+2\left( b+c \right)}+\frac{\sqrt{ca}}{b+2\left( c+a \right)}+\frac{\sqrt{ab}}{c+2\left( a+b \right)}.\] The following inequality is also true. Let $a,b,c$ be positive real numbers . Prove that\[\frac{\sqrt{bc}}{2a+b+c }+\frac{\sqrt{ca}}{2b+c+a }+\frac{\sqrt{ab}}{2c+a+b }\]\[\le \frac{a}{2a+ b+c}+\frac{b}{2b+ c+a }+\frac{c}{2c+ a+b }\le \frac{3}{4}.\]

Let $a,b,c>0$. Prove that $\frac{a-\sqrt{6bc}}{2a+2b+3c}+\frac{2b-\sqrt{3ca}}{4b+3c+a}+\frac{3c-\sqrt{2ab}}{6c+a+2b}\ge 0.$

ionbursuc wrote: The following inequality is true. Let $a,b,c\in \left( 0,\infty \right)$.Prove that $\frac{a-\sqrt{bc}}{2a+b+c}+\frac{b-\sqrt{ca}}{2b+c+a}+\frac{c-\sqrt{ab}}{2c+a+b}\le \frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}+{{\left( \sqrt{b}-\sqrt{c} \right)}^{2}}+{{\left( \sqrt{c}-\sqrt{a} \right)}^{2}}}{2\left( a+b+c \right)}.$ \[\Leftrightarrow \sum \frac{a-\sqrt{bc}}{2a+b+c} \le \sum \frac{a-\sqrt{bc}}{a+b+c}\\\\\Leftrightarrow \sum \frac{a(a-\sqrt{bc})}{(a+b+c)(2a+b+c)} \ge 0 \\\\\Leftrightarrow \sum \frac{a^2}{2a+b+c} \ge \sum \frac{a\sqrt{bc}}{2a+b+c}\] by Caushy we have \[\sum \frac{a^2}{2a+b+c} \ge \frac{(\sum a)^2}{\sum (2a+b+c)}=\frac{a+b+c}{4} \] by AM-GM we have \[ \sum \frac{a\sqrt{bc}}{2a+b+c} \le \sum \frac{a\sqrt{bc}}{2(a+\sqrt{bc})} \le \sum \frac{a+\sqrt{bc}}{8} \le \frac{a+b+c}{4}\] then \[\sum \frac{a^2}{2a+b+c} \ge \frac{a+b+c}{4} \ge \sum \frac{a\sqrt{bc}}{2a+b+c}\]

elmrini wrote: ionbursuc wrote: The following inequality is true. Let $a,b,c\in \left( 0,\infty \right)$.Prove that $\frac{a-\sqrt{bc}}{2a+b+c}+\frac{b-\sqrt{ca}}{2b+c+a}+\frac{c-\sqrt{ab}}{2c+a+b}$ $\le \frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}+{{\left( \sqrt{b}-\sqrt{c} \right)}^{2}}+{{\left( \sqrt{c}-\sqrt{a} \right)}^{2}}}{2\left( a+b+c \right)}.$

$ \sum a\ge \sum \sqrt{bc}+(a+b+c)\sum \frac{a-\sqrt{bc}}{2a+b+c}.$

arqady wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: \[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\] Generalization Let ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{\beta }_{1}},{{\beta }_{2}},{{\beta }_{3}},a,b,c>0$. Prove that $\frac{\left( {{\alpha }_{1}}+{{\beta }_{1}} \right)a-2\sqrt{{{\beta }_{2}}{{\alpha }_{3}}}\sqrt{bc}}{\left( {{\alpha }_{1}}+{{\beta }_{1}} \right)a+{{\beta }_{2}}b+{{\alpha }_{3}}c}+\frac{\left( {{\alpha }_{2}}+{{\beta }_{2}} \right)b-2\sqrt{{{\beta }_{3}}{{\alpha }_{1}}}\sqrt{ca}}{\left( {{\alpha }_{2}}+{{\beta }_{2}} \right)b+{{\beta }_{3}}c+{{\alpha }_{1}}a}+\frac{\left( {{\alpha }_{3}}+{{\beta }_{3}} \right)c-2\sqrt{{{\beta }_{1}}{{\alpha }_{2}}}\sqrt{ab}}{\left( {{\alpha }_{3}}+{{\beta }_{3}} \right)c+{{\beta }_{1}}a+{{\alpha }_{2}}b}\ge 0$

The following inequality is also true. Let $a,b,c,d$ be positive real numbers . Prove that\[{{\frac{a-b}{\sqrt{b+c}}+\frac{b-c}{\sqrt{c+d}}+\frac{c-d}{\sqrt{d+a}}}}+\frac{d-a}{\sqrt{a+b}}\ge 0.\]

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=57775

pcoRO wrote: Let $a,b,c\in \left( 0,\infty \right)$.Prove the inequality $\frac{a-\sqrt{bc}}{a+2\left( b+c \right)}+\frac{b-\sqrt{ca}}{b+2\left( c+a \right)}+\frac{c-\sqrt{ab}}{c+2\left( a+b \right)}\ge 0.$ The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that:\[\frac{a-\sqrt{bc}}{a+2 b+3c}+\frac{b-\sqrt{ca}}{b+2c+3a}+\frac{c-\sqrt{ab}}{c+2a+3b}\ge 0.\]

\[\Longleftrightarrow 3\ge \sum\limits_{cyc}\frac{2b+3c+\sqrt{bc}}{a+2b+3c}\] but $2\sqrt{bc}\le b+c$, so we need to prove that \[6\ge \sum\limits_{cyc}\frac{5b+7c}{a+2b+3c}\Longleftrightarrow6\ge \sum\limits_{cyc}\frac{3(a+2b+3c)-(b+2c+3a)}{a+2b+3c}\Longleftrightarrow\]\[\sum\limits_{cyc}\frac{b+2c+3a}{a+2b+3c}\ge 3\] which is obviously true with AM-GM.

arqady wrote:

The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: \[\frac{a-\sqrt{bc}}{2a+b+c }+\frac{b-\sqrt{ca}}{2b+ c+a}+\frac{c-\sqrt{ab}}{2c+a+b }\ge 0\] Show that for $a,b,c > 0$ the following inequality holds: $$\frac{\sqrt{ab}}{a+b+2c}+\frac{\sqrt{bc}}{b+c+2a}+\frac{\sqrt{ca}}{c+a+2b} \le \frac {3}{4}$$