Ok, i post my solution First,we put $f(a,b,c)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-(a^2+b^2+c^2)$ WLOG $a \leq b \leq c$.We will prove that $f(a,b,c) \geq f(\frac{a+b}{2},\frac{a+b}{2},c)$(1) In fact:(1) $\Longleftrightarrow \frac{1}{a^2}+\frac{1}{b^2}-\frac{8}{(a+b)^2}+\frac{(a+b)^2}{2}-(a^2+b^2) \geq 0$ $\Longleftrightarrow (a-b)^2(\frac{(a+b)^2+2ab}{a^2b^2(a+b)^2}-\frac{1}{2}) \geq 0$ (2) Because $a \leq b \leq c \Longrightarrow a+b \leq 2 \Longrightarrow ab \leq 1$ So that:$a^2b^2(a+b)^2 <2(a+b)^2+4ab \Longrightarrow \frac{(a+b)^2+2ab}{a^2b^2(a+b)^2}-\frac{1}{2} >0 \Longrightarrow$(2) true Hence $f(a,b,c) \geq f(\frac{a+b}{2},\frac{a+b}{2},c)$ (when $a \leq b \leq c$) From this,we need only to prove the case: Give $t,c >0$ satisfy $2t+c=3$.Prove that: $\frac{2}{t^2}+\frac{1}{c^2} \geq 2t^2+c^2$(3) Solution: (3) $\Longleftrightarrow \frac{2}{t^2}+\frac{1}{(3-2t)^2} \geq 2t^2+(3-2t)^2$ $\Longleftrightarrow (t-1)^2(12t^4-36t^3+33t^2-6t-9) \leq 0$ We're easy to check that $12t^4-36t^3+33t^2-6t-9 \leq 0$ when $t \in (0,\frac{3}{2})$ So that the problem is proved

hi, i`m new here the ineq. is equivalent to : $(a+b+c)^{4}(\sum a^{2}b^{2})\geq 81\sum a^{4}b^{2}c^{2}$. we calculate the left side and then AM-GM and Muirhead (or AM-GM only) for the resulting sums, grouped convenably. it is a lot to write, but it is ok

$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2\Leftrightarrow\sum_{cyc}\left(\frac{1}{a^2}-a^2+4(a-1)\right)\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(a-1)^2(1+2a-a^2)}{a^2}\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(2a-b-c)^2((a+b+c)^2+6a(a+b+c)-9a^2)}{a^2}\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(2a-b-c)^2(-2a^2+b^2+c^2+8ab+8ac+2bc)}{a^2}\geq0.$ Let $\frac{-2a^2+b^2+c^2+8ab+8ac+2bc}{a^2}=P_a,$ $\frac{-2b^2+a^2+c^2+8bc+8ab+2ac}{b^2}=P_b,$ $\frac{-2c^2+a^2+b^2+8ac+8bc+2ab}{c^2}=P_c.$ Hance, remain to prove that $\sum_{cyc}(2a-b-c)^2P_a\geq0.$ But $\sum_{cyc}(2a-b-c)^2P_a=\sum_{cyc}(a-b+a-c)^2P_a=$ $=2\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(a-b)^2P_a+\sum_{cyc}(a-c)^2P_a=$ $=2\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(a-b)(a-c+c-b)P_a+\sum(c-a)(c-b+b-a)P_a=$ $=4\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(b-a)(b-c)P_a+\sum_{cyc}(c-a)(c-b)P_a=$ $=\sum_{cyc}(a-b)(a-c)(4P_a+P_b+P_c).$ Let $a\geq b\geq c.$ Then $4P_c+P_a+P_b\geq4P_b+P_a+P_c\Leftrightarrow P_c\geq P_b.$ This is obviously true. And $4P_a+P_b+P_c=-12+\frac{4b^2}{a^2}+\frac{a^2}{b^2}+\frac{4c^2}{a^2}+\frac{a^2}{c^2}+\frac{32b}{a}+\frac{8a}{b}+...\geq0.$ Similarly $4P_b+P_a+P_c\geq0$ and $4P_c+P_a+P_b\geq0.$ And our inequality is proved.

maky wrote: the ineq. is equivalent to : $(a+b+c)^{4}(\sum a^{2}b^{2})\geq 81\sum a^{4}b^{2}c^{2}$. we calculate the left side and then AM-GM and Muirhead (or AM-GM only) for the resulting sums, grouped convenably. it is a lot to write, but it is ok $(a+b+c)^4(a^2b^2+a^2c^2+b^2c^2)\geq81(a^2+b^2+c^2)a^2b^2c^2\Leftrightarrow$ $\Leftrightarrow\sum_{sym}(a^6b^2+4a^5b^3+4a^5b^2c+3a^4b^4+12a^4b^3c+10a^3b^3c^2-34a^4b^2c^2)\geq0.$ AM-GM only.

I got that the inequality is equivalent to $(6,2,0)+4(5,3,0)+4(5,2,1)+3(4,4,0)+12(4,3,1)+10(3,3,2) \geq 34(4,2,2)$. (arqady's post above seems to confirm my result) How do I use AM-GM only? I couldn't cancel the last of the ten $(3,3,2)$ with AM-GM, I have to use Schur to complete the solution: $(4,4,0)+(3,3,2) \geq 2 (4,3,1)$. By the way, why did they give an inequality which can be killed by dumb455-ing??? I find this just plain useless. Anyone who knows this "tehnique" (you can hardly call it a tehnique, but I couldn't find another word) should be able to do it. Also, who proposed this?

perfect_radio wrote: How do I use AM-GM only? $a^6b^2+a^6c^2+4a^5b^3+4a^5c^3+4a^5b^2c+4a^5c^2b+3a^4b^4+3a^4c^4+$ $+12a^4b^3c+12a^4c^3b+10a^3b^3c^2+10a^3c^3b^2\geq68\sqrt[68]{a^{272}b^{136}c^{136}}=68a^4b^2c^2.$

The form of the inequality is nice. Then, I think you can find at least two nice solutions to this inequality.

Give at least one mr Cirtoaje . BTW, my solution is similar to Arqady's one . Arquady thank you for teaching me the trick with the first derivative to the factorisation.

I think that this one might interest you too : http://www.mathlinks.ro/Forum/viewtopic.php?t=70815 .

A nice method: Use the first two lines from Arqady's solution, then considere two cases: Case $c\leq b \leq a \leq 1+\sqrt{3}$; Case $a>1+\sqrt{3}$, when $b+c<1/2$, $bc<1/16$ and $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}>\frac 2{bc}>32>(a+b+c)^2>a^2+b^2+c^2$.

Very nice Thank you mr Vasile .You are the constructor of this nice problem ??

ciprian wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] In my solution I have proved that \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] <=> \[ (b-c)^2S_a+(a-c)^2S_b+(a-b)^2S_c\geq 0 \] wher $S_a=\frac{2}{b^2}+\frac{2}{c^2}+\frac{6}{bc}-\frac{1}{a^2}-\frac{27}{(a+b+c)^2}$ $S_b$ and $S_c$ definding similary. And with leting $a\geq b\geq c$ I proved that $S_a,S_b\geq 0$ and $S_b+S_c\geq 0$,and how we know from sos the ineq is done .

Vasc wrote: A nice method: Use the first two lines from Arqady's solution, then considere two cases: Case $c\leq b \leq a \leq 1+\sqrt{3}$; Case $a>1+\sqrt{3}$, when $b+c<1/2$, $bc<1/16$ and $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}>\frac 2{bc}>32>(a+b+c)^2>a^2+b^2+c^2$. Vasc, you meant to: Case $c\leq b \leq a \leq 1+\sqrt{2}$; Case $a>1+\sqrt{2}$, when $b+c<3/5$, $bc<9/100$ and $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}\geq\frac 2{bc}>9=(a+b+c)^2>a^2+b^2+c^2$. OK?

Now that you've literally killed the problem, what do you think about this: Quote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. For what $t \in \mathbb R$ is it true that we always have \[ \frac 1{a^t}+\frac 1{b^t}+\frac 1{c^t} \geq a^t + b^t + c^t \, ? \]

it is easy to prove when $t \in [0,1]$ but...

arqady wrote: Vasc, you meant to: Case $c\leq b \leq a \leq 1+\sqrt{2}$; Case $a>1+\sqrt{2}$, when $b+c<3/5$, $bc<9/100$ and $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}\geq\frac 2{bc}>9=(a+b+c)^2>a^2+b^2+c^2$. OK? You are right Arqady. It must $\sqrt 2$, not $\sqrt 3$. Now this solution is OK. But can someone find the second nice solution (actually, the first nice solution) ?

Since $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 1{ab}+\frac 1{bc} +\frac 1{ca}$, it suffices to show that $\frac 1{ab}+\frac 1{bc} +\frac 1{ca} \geq a^2+b^2+c^2$. This is my third solution.

This problem is added to my book recently. My solution is to used Chebyshev only, I think it's very simple and nice enough. We need to prove that \[ \sum_{sys} a^2b^2(1-c^4) \ge 0 \] First case, if $a \ge b \ge c$ and $ab,bc,ca \le 2$ then \[ a^2b^2(1+c+c^2+c^3) \ge a^2c^2(1+b+b^2+b^3) \ge b^2c^2(1+a+a^2+a^3) \] So by Chebyshev, we have done (notice that $a+b+c=3$). Otherwise, suppose that $ab \ge 2$, so $a+b \ge 2\sqrt{2}$ and $c \le \frac{1}{3}$, thus \[ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} >9>a^2+b^2+c^2. \] So the problem is completely solved.

Vasc wrote: Since $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 1{ab}+\frac 1{bc} +\frac 1{ca}$, it suffices to show that $\frac 1{ab}+\frac 1{bc} +\frac 1{ca} \geq a^2+b^2+c^2$. This is my third solution. Are you meant to $\frac 1{ab}+\frac 1{bc} +\frac 1{ca} \geq a^2+b^2+c^2\Leftrightarrow(a+b+c)^5\geq81(a^2+b^2+c^2)abc$ and to use a follow inequality http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%2A2A&t=62923 ?

Vasc wrote: Since $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 1{ab}+\frac 1{bc} +\frac 1{ca}$, it suffices to show that $\frac 1{ab}+\frac 1{bc} +\frac 1{ca} \geq a^2+b^2+c^2$. This is my third solution. Applying AM-GM $\sum_{cyc} \frac{1}{a^2}\geq \sum_{cyc} \frac{1}{ab}=\frac{a+b+c}{abc}=\frac{3}{abc}=\frac{27}{3abc(a+b+c)}\geq \frac{27}{(ab+bc+ca)^2}$ We need to show that $\frac{27}{(ab+bc+ca)^2}\geq a^2+b^2+c^2\Leftrightarrow (ab+bc+ca)^2(a^2+b^2+c^2)\leq 27$ By AM-GM: $(ab+bc+ca)^2(a^2+b^2+c^2)=(\sum_{cyc} ab)(\sum_{cyc} ab)(\sum_{cyc} a^2)\leq \left [ \frac{(a+b+c)^2}{3} \right ]^3$ The equality holds at $a=b=c=1$.

Solution taken out.

Sohil_Doshi wrote: By Chebyshev's inequality, $3(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2$ Now we essentially have to prove that $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2 \geq 3(a^2+b^2+c^2)$ $(\frac{ab+bc+ca}{abc})^2 \geq 3(a^2+b^2+c^2)$ By AM GM $ab+bc+ca \geq 3(abc)^{2/3}$ $(ab+bc+ca)^3 \geq 27(abc)^2$ $(ab+bc+ca)^3 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$ $27(abc)^2 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$ $(a^2+b^2+c^2)(ab+bc+ca) \leq 9$ By Chebyshev's inequality, $(a^2+b^2+c^2)(bc+ca+ab) \leq 3abc(a+b+c)$ $(a^2+b^2+c^2)(bc+ca+ab) \leq 9abc$ So if we can prove that $abc \leq 1$ then we are done By AM GM, $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$ Since $a+b+c=3$ this implies that $abc \leq 1$. $\blacksquare$ Your application of Chebyshev's inequality is not right.

bel.jad5 wrote: Sohil_Doshi wrote: By Chebyshev's inequality, $3(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2$ Now we essentially have to prove that $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2 \geq 3(a^2+b^2+c^2)$ $(\frac{ab+bc+ca}{abc})^2 \geq 3(a^2+b^2+c^2)$ By AM GM $ab+bc+ca \geq 3(abc)^{2/3}$ $(ab+bc+ca)^3 \geq 27(abc)^2$ $(ab+bc+ca)^3 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$ $27(abc)^2 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$ $(a^2+b^2+c^2)(ab+bc+ca) \leq 9$ By Chebyshev's inequality, $(a^2+b^2+c^2)(bc+ca+ab) \leq 3abc(a+b+c)$ $(a^2+b^2+c^2)(bc+ca+ab) \leq 9abc$ So if we can prove that $abc \leq 1$ then we are done By AM GM, $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$ Since $a+b+c=3$ this implies that $abc \leq 1$. $\blacksquare$ Your application of Chebyshev's inequality is not right. Which application and I am not able to see why it is not right. How is it not right?

We can use AM-GM to get \[\dfrac 1{a^2}+\dfrac 1{b^2}+\dfrac 1{c^2}\geq\dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}=\dfrac{a+b+c}{abc}=\dfrac 3{abc}\]However, we note that using the inequality \[(x+y+z)^2\geq 3(xy+yz+xz)\]for $x=bc,y=ac,z=ab$, we get that \[(ab+bc+ca)^2\geq 3(a^2bc+ab^2c+abc^2)=3abc(a+b+c)=9abc\]Thus, we can substitute to get \[\dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}\geq\dfrac{3}{abc}\geq\dfrac{27}{(ab+bc+ca)^2}\]However, by AM-GM, we have \[27(ab+bc+ca)(ab+bc+ca)(a^2+b^2+c^2)\leq (a^2+b^2+c^2+ab+bc+ca+ab+bc+ca)^3=((a+b+c)^2)^3=(a+b+c)^6\]Thus, we get that \[\dfrac 1{a^2}+\dfrac 1{b^2}+\dfrac 1{c^2}\geq \dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}\geq\dfrac{27}{(ab+bc+ca)^2}\geq\dfrac{729(a^2+b^2+c^2)}{(a+b+c)^6}=a^2+b^2+c^2\]where we used $a+b+c=3$.

Sohil_Doshi wrote: $(a^2+b^2+c^2)(bc+ca+ab) \leq 3abc(a+b+c)$ This step is wrong - inequality is the other way simply by AM-GM \[bc+ca+ab\geq 3(abc)^{2/3}\]and \[\dfrac 23a^2+\dfrac 16b^2+\dfrac 16c^2\geq 6\sqrt[6]{\dfrac{a^8b^2c^2}{6^{6}}}=(abc)^{1/3}a\tag{*}\]because adding the symmetric versions gives \[a^2+b^2+c^2\geq (abc)^{1/3}(a+b+c)\]Now, multiplying the two inequalities yields the conclusion.

Thank you for the explanation. I have edited that part of the proof and used a different method of proving the inequality $(a^2+b^2+c^2)(ab+bc+ca) \leq 9$.

Sohil_Doshi wrote: Thank you for the explanation. I have edited that part of the proof and used a different method of proving the inequality $(a^2+b^2+c^2)(ab+bc+ca) \leq 9$. The inequality mentioned is incorrect. Try the counter-example $(x,y,z)=(\frac 32,\frac 34,\frac 34)$.

ciprian wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] Let $q=ab+bc+ca$ , $ r=abc$, $ \rightarrow r \le 1$ , $ q^2 \ge 9r$ Write the inequality in this form $$LHS-RHS=\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) -9=\frac{q^2-6r}{r^2}+2q-9 \ge \frac{q^2-6(\frac{q^2}{9})}{ (\frac{q^2}{9})^2}+2q-9= \frac{81(q-3)^2(2q+3)}{q^2}\ge 0$$

ciprian wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] $$\text{LHS-RHS}=\,{\frac {[ \left( ab+ca+bc \right) ^{2}+3\,abc]\cdot {[{{a}^{2} \left( b -c \right) ^{2}+{b}^{2} \left( c-a \right) ^{2}+{c}^{2} \left( a-b \right) ^{2}}]}}{2{a}^{2}{b}^{2}{c}^{2} \left( ab+ca+bc \right) ^{2}}}+{\frac { \left( 2\,ab+2\,ca+2\,bc+3 \right) \left( ab+ca+bc-3 \right) ^{2}}{ \left( ab+ca+bc \right) ^{2}}} \geqq 0$$I have another but ugly

ciprian wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] Equivalent to prove $$ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) \geq 9 $$ We write $$ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) \ge \frac 1{ab}+\frac 1{bc}+\frac 1{ca}+2(ab+bc+ca) =\frac 3{abc}+2(ab+bc+ca) \ge \frac {27}{(ab+bc+ca)^2}+2(ab+bc+ca) \ge 3\sqrt[3]{ \frac{27}{(ab+bc+ca)^2} \cdot (ab+bc+ca)^2 }=9 $$

ciprian wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \] I like this one. So our ineq is equivalent to $$\sum_{\text{cyc}} \frac{1}{a^2}-a^2 \ge 0$$Now remember that $x+\frac{1}{x} \ge 2$ so $\frac{1}{x^2}-x^2 \ge 2 \left(\frac{1}{x}-x \right)$ and now let $f(x)=\frac{1}{x}-x$, then $f''(x)=\frac{2}{x^3}>0$ so $f$ is convex so by jensen $f(a)+f(b)+f(c) \ge 0$ and now replacing all of this stuff we get $$\sum_{\text{cyc}} \frac{1}{a^2}-a^2 \ge 2 \sum_{\text{cyc}} \frac{1}{a}-a \ge 0$$Thus we are done

Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{32}{81}(a^2+4b^2+c^2)$$h

Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that $$\frac {a}{b^3}+\frac {b^3}{c^5}+\frac {c^5}{a} \geq a^2+b^2+c^2$$$$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 3{2} + \frac 1{2} (a^3+b^3+c^3)$$$$\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\ge a^3+b^3+c^3$$Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$\frac {a}{b^2}+\frac {b^3}{c^5}+\frac {c^5}{a} \geq a^2+b^2+c^2$$$$\sqrt{\frac {a(1+b)}{b(1+c)}}+\sqrt{\frac {b(1+c)}{c(1+a)}}+\sqrt{\frac {c(1+a)}{a(1+b)}}\leq a+b+c$$$$a^3+b^3+c^3-a^2b-b^2c-c^2a \geq \frac {479}{100} (a+b+c-3)$$$$a^3+b^3+c^3-a^2b-b^2c-c^2a \geq \frac {84}{25} (ab+bc+ca-3)$$Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3$. Prove that $$\frac {1+a}{1+b^3}+\frac {1+b^3}{1+c^5}+\frac {1+c^5}{1+a} \geq a+b+c$$$$\frac {1+a}{1+b^3}+\frac {1+b}{1+c^3}+\frac {1+c}{1+a^3} \geq a+b+c$$ sqing wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^3}+\frac 1{b^3}+\frac 1{c^3} \geq a^3+b^3+c^3. \]How to prove? h h Let $a,b,c>0, \ a+b+c+abc=4$. Prove that $$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge a^3+b^3+c^3$$

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ij realized this problem was posted before I was born