goodar2006 wrote: For positive reals $a,b$ and $c$ with $ab+bc+ca=1$ show that: $\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}}$ My solution: Define $\sqrt {a} =x , \sqrt{b}= y , \sqrt{c} =z$ We know $\sum {x^2y^2}=1$ and we want to prove that $\sum {\frac{x^3}{y^2z^2}} \geq \ \sqrt {3} \sum x$ Now use Holder inequality : $\sum {\frac{x^3}{y^2z^2}} (\sum yz) (\sum yz) \geq (\sum x)^3 $ We know by CS that $\sum xy \le \sqrt {3}$ and $(\sum x)^2 \geq 3 (\sum xy) $

I hope there's no mistake (I haven't done any inequalities in a long time!): WLOG $a\ge b\ge c$, so by Chebyshev's inequality, $\left(\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)(\sqrt{bc}+\sqrt{ca}+\sqrt{ab})\ge 3\left(\frac{a\sqrt{a}}{\sqrt{bc}}+\frac{b\sqrt{b}}{\sqrt{ca}}+\frac{c\sqrt{c}}{\sqrt{ab}}\right)$ Now $\frac{a\sqrt{a}}{\sqrt{bc}}+\frac{b\sqrt{b}}{\sqrt{ca}}+\frac{c\sqrt{c}}{\sqrt{ab}}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}$ by Muirhead as $\left[\frac{3}{2},\frac{-1}{2},\frac{-1}{2}\right]\ge\left[\frac12,0,0\right]$. Alternatively $\sum \left(2\frac{a\sqrt{a}}{\sqrt{bc}}+\frac{b\sqrt{b}}{\sqrt{ca}}+\frac{c\sqrt{c}}{\sqrt{ab}}\right)\ge\sum4\sqrt{a}$ by AM-GM. So $\left(\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)(\sqrt{bc}+\sqrt{ca}+\sqrt{ab})\ge 3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$ So $\frac{\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}}{\sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})}\ge\frac{\sqrt{3}}{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}$ Thus we want to prove $\frac{\sqrt{3}}{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}\ge 1$. But this follows from cauchy: $(ab+bc+ca)(1+1+1)\ge (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^2$

Muirhead inequality for positive numbers are

matth wrote: Muirhead inequality for positive numbers are Should we read it as a question? (Of course no punctuation marks....) Anyway, may solution: After making the inequality homogenous, we have to prove: $(\sum a^2\sqrt{a})^2(\sum ab)\ge 3a^2b^2c^2(\sum \sqrt{a})^2$ Which is obvious, since by Muirheads inequality, $(\sum a^5)(\sum ab)\ge 3a^2b^2c^2(\sum a)$ and $(\sum a^2b^2\sqrt{ab})(\sum ab)\ge 3a^2b^2c^2(\sum \sqrt{a})$.

goodar2006 wrote: For positive reals $a,b$ and $c$ with $ab+bc+ca=1$ show that: $\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}}$ It is very amazing in Iran TST 2012 have a very easy problem

dm_edogawasonichi wrote: It is very amazing in Iran TST 2012 have a very easy problem Yes , It seem this is the easiest one until now . There are many other solutions , here is one that I wrote during the exam : $\sum \frac{a\sqrt {a}}{bc} = \sum \frac{a^3}{a^{\frac{3}{2}}bc} \\ \\ \\ \geq \frac{(\sum a^{\frac{3}{2}} )^2}{abc(\sum a^{\frac{1}{2}})} = \frac{(\sum a^{\frac{3}{2}} )^2}{\sum a^{\frac{1}{2}}} . \left ( \sum \frac{1}{a} \right ) . \left ( \sum ab \right ) \\ \\ \\ \geq \frac{(\sum a^{\frac{3}{2}} )^2}{\sum a^{\frac{1}{2}}} . \left (\sum a^{\frac{1}{2}} \right )^2 \\ \\ \\ = \left (\sum a^{\frac{3}{2}} \right )^2 . \left (\sum a^{\frac{1}{2}} \right ) $ And by Holder inequality we have $(1+1+1) \left (\sum a^{\frac{3}{2}} \right )^2 \geq \left (\sum a \right )^3 \geq \left (\sqrt{3 \sum ab} \right )^3 =3\sqrt{3}$

goodar2006 wrote: For positive reals $a,b$ and $c$ with $ab+bc+ca=1$ show that: $\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}}$ Surprisingly very easy problem for Iran TST.. Here is my proof: WLOG $a\ge b\ge c$ so using Chebishev's and Cauchy-Schwarz inequality , we have: $ \sum \frac{a\sqrt{a}}{bc} \ge \frac{1}{3} \left( \sum \frac{a}{bc} \right) \left( \sum \sqrt{a} \right)\ \ge \frac{(a+b+c)^2}{9abc}\sum \sqrt{a}$ Hence, it remains to prove: $(a+b+c)^2 \ge 9\sqrt{3}abc$, After homogenization it's equal to: $(a+b+c)^2 \sqrt{ab+ac+bc} \ge 9\sqrt{3}abc$ , and this follows from am-gm. Equality holds for $a=b=c= \frac{1}{\sqrt{3}}$. $\blacksquare$

goodar2006 wrote: For positive reals $a,b$ and $c$ with $ab+bc+ca=1$ show that: $\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}}$ We have to prove: ${\sqrt{a}+\sqrt{b}+\sqrt{c}\le \left( \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)\sqrt{\frac{ab+bc+ca}{3}}}$ Using AM-QM and Cauchy-Schwarz and AM-GM inequality we have: $\left( \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)\sqrt{\frac{ab+bc+ca}{3}}\geq \left( \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}\\ =\left( \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)(\sqrt{a}bc+\sqrt{b}ca+\sqrt{c}ab)\frac{1}{3\sqrt{abc}}\geq\frac{(a+b+c)^2}{3\sqrt{abc}}\\ \geq\frac{ab+bc+ca}{\sqrt{abc}}\geq\frac{\sqrt{a}bc+\sqrt{b}ca+\sqrt{c}ab}{\sqrt{abc}}={\sqrt{a}+\sqrt{b}+\sqrt{c}}$

matth wrote: Muirhead inequality for positive numbers are Do you mean to say Muirhead is only for positive exponents? Actually, Muirhead works for positive/negative/zero exponents. Anyway, you could just multiply the whole inequality by $abc$ to get $\left[\frac{5}{2},\frac{1}{2},\frac{1}{2}\right]\ge\left[\frac32,1,1\right]$ which is true, then divide by $abc$ again

Though unnecessary, here's another proof: Using the Cauchy-Schwarz inequality, we have \[(1): \ \ \left(\frac{a\sqrt a}{bc}+\frac{b\sqrt b}{ca}+\frac{c\sqrt c}{ab}\right)\left(\sqrt a+\sqrt b+\sqrt c\right)\geq \left(\frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ca}}+\frac{c}{\sqrt{ab}}\right)^2;\] And \[(2): \ \ \left(\frac{a}{\sqrt{bc}}+\frac{b}{\sqrt{ca}}+\frac{c}{\sqrt{ab}}\right)\left(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right)\geq (\sqrt a+\sqrt b +\sqrt c)^2\] Note that performing $(1)\times(2)^2,$ we get; \[\left(\frac{a\sqrt a}{bc}+\frac{b\sqrt b}{ca}+\frac{c\sqrt c}{ab}\right)\left(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right)^2\geq (\sqrt a+\sqrt b +\sqrt c)^3\] Thus it is sufficient to check that \[\sqrt a+\sqrt b+\sqrt c\geq \sqrt[4]{3}\left(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right);\] Now, note that $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq \sqrt{3(ab+bc+ca)}=\sqrt 3;$ and therefore it is sufficient to check that \[\left(\sqrt a+\sqrt b+\sqrt c\right)^2\geq 3\left(\sqrt{bc}+\sqrt{ca}+\sqrt{ab}\right);\] Which is an obvious application of the AM-GM inequality. We are done. Equality holds iff $a=b=c=\sqrt{3}/3.\Box$

Lol, am I so bad in inequalities ? I did it pretty quickly, but first one from first day was in my opinion way easier. From $\sum \frac{x_i}{y_i} \geq \frac{(\sum x_i)^2}{\sum x_iy_i}$ (this is one of the CS varations) we have $\sum \frac{a \sqrt{a}}{bc} \geq \frac{( \sum a \sqrt{a})^2}{abc \sum \sqrt{a}}$ We will now prove $\frac{ (\sum a \sqrt{a})^2}{abc \sum \sqrt{a}} \geq \sqrt{3} ( \sum \sqrt{a})$. It's equivalent to $(\sum a \sqrt{a})^4 (ab+bc+ca) \geq 3 ( \sum \sqrt{a})^4 a^2b^2c^2$ but $\frac{ \sum a \sqrt{a} }{ \sum \sqrt{a} } \geq \frac{a+b+c}{3}$ so $\frac{(\sum a \sqrt{a})^4}{ ( \sum \sqrt{a})^4 } \frac{ab+bc+ca}{3} \geq (\frac{a+b+c}{3})^4 \frac{ab+bc+ca}{3} \geq a^2b^2c^2$ and we're done.

$ a=x^2$ etc. then $(\sum{x^2}) ^4 \ge 9$ .also , $(\sum{x^2})^2 \ge 3$ $1=\sum{x^2y^2} \ge xyz(z+x+y)$ . $([\sum{xy}]^2) \le 3$ hence $ (\sum{x^3/y^2z^2}) \ge \frac{[\sum{x^2}]^2}{xyz(\sum{xy})} \ge (x+y+z).3^{1/2}$

Easy one! $\left(\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)^2\left(a^3b^2c^2+b^3c^2a^2+c^3a^2b^2\right)\ge (a^2+b^2+c^2)^3$ $\implies \left(\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}\right)\ge \sqrt{\frac{(a^2+b^2+c^2)^3}{\left(a^3b^2c^2+b^3c^2a^2+c^3a^2b^2\right)}}$ But $\sqrt{\frac{(a^2+b^2+c^2)^3}{\left(a^3b^2c^2+b^3c^2a^2+c^3a^2b^2\right)}}=\sqrt{\frac{(a^2+b^2+c^2)^3}{a^2b^2c^2(a+b+c)\right}}$ Hence it remains to prove $\sqrt{\frac{(a^2+b^2+c^2)^3}{a^2b^2c^2(a+b+c)\right}}\ge \sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})}$ $\implies \frac{(a^2+b^2+c^2)^3}{a^2b^2c^2(a+b+c)\right}\ge 3(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ We have $1=(ab+bc+ca)^2\ge3 abc(a+b+c)$ $\implies \frac{(a^2+b^2+c^2)^3}{a^2b^2c^2(a+b+c)\right}\ge \frac{3(a^2+b^2+c^2)^3}{abc}$ $ \frac{3(a^2+b^2+c^2)^3}{abc}\ge 3(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ $\implies (a^2+b^2+c^2)^3\ge abc(\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ But we have $3(a+b+c)\ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ Hence, it remains to prove $(a^2+b^2+c^2)^3\ge 3abc(a+b+c)$ We have $(ab+bc+ca)^2\ge 3abc(a+b+c)$ Remains to prove $ (a^2+b^2+c^2)^3\ge (ab+bc+ca)^2$ Which follows from $a^2+b^2+c^2\ge ab+bc+ca\implies (a^2+b^2+c^2)^3\ge 1=(ab+bc+ca)^2\blacksquare$

Absolutely trivial: First,we prove that 1/a + 1/b + 1/c >= 3*sqrt(3) 1/a + 1/b + 1/c = 1/abc , so it remains to prove that abc<=1/3*sqrt(3),whic is obvios by AM-GM Now,put sqrt(a)=x,sqrt(b)=y,sqrt(c)=z We will prove this: (1/x^2 + 1/y^2 + 1/z^2)(x+y+z)<=3*(x^3/y^2*z^2+y^3/x^2*z^2+z^3/x^2*y^2) After multipling the LHS we have Muirheads inequality for [3,-2,-2]>=[-1,0,0] [3,-2,-2]>=[1,-2,0], so we are finished.(The Murhiead can easy be avoided just doing AM-GM,but I am week to write it)

goodar2006 wrote: For positive reals $a,b$ and $c$ with $ab+bc+ca=1$, show that \[\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}.}\] Proposed by Morteza Saghafian $\frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}=\frac{a^2}{\sqrt{a} bc}+\frac{b^2}{a\sqrt{b}c}+\frac{c^2}{ab\sqrt{c}}\ge\frac{(a+b+c)^2}{\sqrt{abc}(\sqrt{bc}+\sqrt{ca}+\sqrt{ab})}$ $\ge\frac{(a+b+c)^2}{\sqrt{3abc}}\ge\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c}).}$ or $\sum \frac{a\sqrt{a}}{bc}+2\cdot 3^{\frac{3}{4}}\sum \sqrt{bc}\ge 3\sqrt{3}\sum \sqrt{a}\ge \sqrt{3}\sum \sqrt{a}+2\cdot 3^{\frac{3}{4}}\sum \sqrt{bc}$ $\Rightarrow \sum\frac{a\sqrt{a}}{bc}\ge \sqrt{3}\sum\sqrt{a}.$

Yes, very easy problem ( 2 min) for IRANTST-2012-EXAM2. Here my solution: we must prove that $ a^{\frac{5}{2}}+b^{\frac{5}{2}}+c^{\frac{5}{2}} \ge \sqrt{3}(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}} $.By chebishev's inequality we have $ 3(a^{\frac{5}{2}}+b^{\frac{5}{2}}+c^{\frac{5}{2}})\ge (a^2+b^2+c^2)(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}) \ge (a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}) \sqrt{3}(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}} $, because by am-gm we have $ a^2+b^2+c^2 \ge 1 $ and $ 3\sqrt{3}abc \le 1 $, so we are done!

Yes, very easy problem ( 2 min) for IRANTST-2012-EXAM2. Here my solution: we must prove that $ a^{\frac{5}{2}}+b^{\frac{5}{2}}+c^{\frac{5}{2}} \ge \sqrt{3}(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}} $.By chebishev's inequality we have $ 3(a^{\frac{5}{2}}+b^{\frac{5}{2}}+c^{\frac{5}{2}})\ge (a^2+b^2+c^2)(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}) \ge (a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}) \sqrt{3}(a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}} $, because by am-gm we have $ a^2+b^2+c^2 \ge 1 $ and $ 3\sqrt{3}abc \le 1 $, so we are done!

Set $x = \sqrt{a}, y = \sqrt{b}, z = \sqrt{c}$ for positive reals $x, y, z.$ Then $x^2y^2 + y^2z^2 + z^2x^2 = 1$, and the inequality is equivalent to \[\sqrt{3}\left(x + y + z\right) \le \frac{x^3}{y^2z^2} + \frac{y^3}{z^2x^2} + \frac{z^3}{x^2y^2}.\] Upon multiplication by $2xyz$, we wish to prove that \[2\sqrt{3}xyz(x + y + z) \le 2\left(\frac{x^4}{yz} + \frac{y^4}{zx} + \frac{z^4}{xy}\right).\] Now we seek to rewrite the entire inequality in terms of $t = xy + yz + zx.$ Note that \[2xyz(x + y + z) = \left(xy + yz + zx\right)^2 - x^2y^2 - y^2z^2-z^2x^2 = t^2 - 1.\] In addition, by AM-GM: \[\sum\limits_{cyc} \frac{x^4}{yz} + \frac{x^4}{yz} + \frac{y^4}{zx} + \frac{y^4}{zx} + \frac{z^4}{xy} \ge \sum\limits_{cyc} 5xy \implies \frac{x^4}{yz} + \frac{y^4}{zx} + \frac{z^4}{xy} \ge t.\] Hence, it suffices to prove that \[\sqrt{3}\left(t^2 - 1\right) \le 2t \iff (t - \sqrt{3})(\sqrt{3}t + 1) \le 0.\] It remains to show that $t \le \sqrt{3}$, but is just Cauchy-Schwarz: \[\left(x^2y^2 + y^2z^2 + z^2x^2\right)\left(1 + 1 + 1\right) \ge \left(xy + yz + zx\right)^2 \implies 3 \ge t^2.\] $\square$