Expanding and multiply each side by 2, we obtain\[2\sum\frac{a^2}{b^2}+2\sum\frac{b}{a}\ge6+2\sum\frac{a}{b}\]Now by AM-GM, $\frac{a^2}{b^2}+\frac{a^2}{b^2}+\frac{b}{a}\ge3\frac{a}{b}$. Now it suffices to prove $\sum\frac{a}{b}+\sum\frac{b}{a}\ge6$, which is obvious btw, I think you'd better post inequality in the inequality forum, rather than algebra(though $\text{Inequality}\in\text{Algebra}$ ) [edit]: (I think)post already moved to inequality by mod

This problem was given in the British Maths Olympiad 2005 as well.

We have from Cauchy Schwarz inequality: $ \left(\frac ab +\frac bc +\frac ca\right)(ab+bc+ca) \geq (a+b+c)^2 $ and $ \left(\frac ab +\frac bc +\frac ca\right)\left(\frac {1}{ab} +\frac {1}{bc} +\frac {1}{ca}\right) \geq \left(\frac 1b +\frac 1c +\frac 1a\right)^2 $ but $ (ab+bc+ca)\left(\frac {1}{ab} +\frac {1}{bc} +\frac{1}{ca}\right)=(a+b+c)\left(\frac 1a +\frac 1b +\frac 1c\right)$ Hence multiplying the above two inequalities we get the desired result.

@ British MO: Indeed, see http://www.mathlinks.ro/Forum/viewtopic.php?t=29253 or http://www.mathlinks.ro/Forum/viewtopic.php?t=25355 . darij

Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[(\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)\] $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2$=$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ and $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$=$3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $\rightarrow $ we want to prove that : $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ $ \geq$ $3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $rightarrow $ we want to prove this inequality: $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+ \frac{b}{a}+\frac{a}{c}+\frac{c}{b}$ $ \geq$ $3+ \frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{b}{a}+\frac{a}{c}+\frac{c}{b} \geq 3$(?) so now we should prove that $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{(\frac{a}{b}+\frac{c}{a}+\frac{b}{c})^2}{3}$(?) and we have $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ $\geq$ $3$(?) $\rightarrow $ $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ and now we r done.

Here is another solution for this inequality By some obvoius calculations the inequality is equivalent with: $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^{2}\geq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$ Anyway, by substitutions $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ the inequality becomes $(x+y+z)^{2}\geq 3+x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ plus condition $xyz=1.$ After all, the inequality is equivalent with this obvious one: $x^{2}+y^{2}+z^{2}+xy+yz+zx\geq 3+x+y+z.$

Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[ (\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c) \] let's take it easy: $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}+ \frac{2a}{c} + \frac{2c}{b} + \frac{2b}{a}$ $\geq$ $1+1+1 +$ $\frac ab$ $+ \frac ac$ $+ \frac ba$ $+ \frac{b}{c}+ \frac{c}{b} + \frac{c}{a}$ now: if we solve this , the problem will solved: $\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0;$ $(\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0) b^2a \longleftrightarrow a^3 + b^3 \geq a^2b + b^2a$

ok is that right?

Put $x=\frac{a}{b}$ & $y,z$ are defined similarly.now multiplying & expanding we have to show $x^2+y^2+z^2+xy+yz+zx\ge 3+x+y+z$.now $xyz=1$ so am-gm gives $\sum_{cyc}xy\ge 3 (*_1)$ now consider $f(t)=t^2-t$ which is convex.so $ f(x)+f(y)+f(z)\ge3f(\frac{x+y+z}{3})$ again $f$ is increasing for $x\ge 0.5$ now am-gm again gives $x+y+z\ge 3$ so $f(\frac{x+y+z}{3})\ge f(1)=0$ which gives $x^2+y^2+z^2\ge x+y+z(*_2)$.adding $(*_1)$ & $(*_2)$ we have the reqd ineq. Sorry for reviving this post but your comments will be helpful

jensen wrote: Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[(\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c)\] $(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2$=$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ and $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$=$3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $\rightarrow $ we want to prove that : $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{b}{a}+2\frac{a}{c}+2\frac{c}{b}$ $ \geq$ $3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}$ $rightarrow $ we want to prove this inequality: $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+ \frac{b}{a}+\frac{a}{c}+\frac{c}{b}$ $ \geq$ $3+ \frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{b}{a}+\frac{a}{c}+\frac{c}{b} \geq 3$(?) so now we should prove that $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ we have: $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{(\frac{a}{b}+\frac{c}{a}+\frac{b}{c})^2}{3}$(?) and we have $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ $\geq$ $3$(?) $\rightarrow $ $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ and now we r done. $ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ $\geq$ $\frac{a}{b}+\frac{c}{a}+\frac{b}{c}$ please check this step.It is yet not done.

pirhoosh wrote: Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[ (\frac ab+\frac bc+\frac ca)^2\geq (a+b+c)(\frac1a+\frac1b+\frac1c) \] let's take it easy: $\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2}+ \frac{2a}{c} + \frac{2c}{b} + \frac{2b}{a}$ $\geq$ $1+1+1 +$ $\frac ab$ $+ \frac ac$ $+ \frac ba$ $+ \frac{b}{c}+ \frac{c}{b} + \frac{c}{a}$ now: if we solve this , the problem will solved: $\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0;$ $(\frac{a^2}{b^2} + \frac{b}{a} - \frac{a}{b} -1 \geq 0) b^2a \longleftrightarrow a^3 + b^3 \geq a^2b + b^2a$ yes the lemma is true and it's a very nice solution

i think we can use muirhead

I don't see any error in Jansen's proof.Please tell me tat, where is the problem ?

hello SMOJ can you show your muirhead solution please?

@SMOJ the ineq is not symmetric.you can't apply muirhead.

In $\triangle ABC$, Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq 3(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})\geq(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]

hey sqing the triangle condition is not necessary: $\implies \sum{\frac{a^2}{b^2}}+\sum{\frac{2a}{c}}\geq \sum{\frac{3a}{c}}$ $\implies \sum{\frac{a^2}{b^2}}\geq \sum{\frac{a}{c}}$ which is $\frac{a^2}{b^2}+\frac{b^2}{c^2}\geq \frac{2a}{c}$ abd ading the similar inequalities... check the triangle condition...

oo.... maybe for the second ineq. it is important....

yes,yes for the second one the triangle condition is indeed important... after the usual Ravi substitution it is coming down to $x^3+y^3+z^2\geq xy^2+yz^2+zx^2$ which is true by re-arrangement

My Solution. By subtracting $RHS$ from $LHS$ we get $-$ $\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)$ $=\sum\dfrac{a^2}{b^2}+2\left(\sum\dfrac{a}{c}\right)-\left[3+\left(\sum\dfrac{a}{b}+\sum \dfrac{a}{c}\right)\right]~.$ $=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}\right)+\sum\dfrac{a}{c}-3~.$ $=\sum\left(\dfrac{a^2}{b^2}-\dfrac{a}{b}+\dfrac{1}{4}\right)+\sum\dfrac{a}{c}-\dfrac{15}{4}~.$ $=\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}.$ It follows from $AM-GM$ that $\sum\dfrac{a}{c}\ge 3~.~(1)$ And by $CS$ we get $-$ $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(\sum \dfrac{a}{b}-\dfrac{3}{2}\right)^2}{3}\xrightarrow{A\ge G}\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2\geq \dfrac{\left(3-\dfrac{3}{2}\right)^2}{3}=\dfrac{3}{4}~.~(2)$ By adding $(1),(2)$ we get $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}\geq 3+\dfrac{3}{4}=\dfrac{15}{4}~.$ Or, $\sum\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2+\sum\dfrac{a}{c}-\dfrac{15}{4}\geq 0~.$ Or, $\left(\sum\dfrac{a}{b}\right)^2-\left(\sum a\right)\left(\sum\dfrac{1}{a}\right)\geq 0$ Or, $\boxed{\left(\sum\dfrac{a}{b}\right)^2\geq \left(\sum a\right)\left(\sum\dfrac{1}{a}\right)}~.~\square$

Prove, that for $a,b,c>0$ \[(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2 \geq\sqrt{3(a^2+b^2+c^2)}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\]

$x,y,z>0$,prove that \[ \left( {\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{x}} \right) ^{2}- \left( x+y+z \right) \left( {\frac {1}{x}}+{\frac {1}{y}}+{\frac {1} {z}} \right) \geq \left( \frac{3}{2}+{\frac {5}{18}}\,\sqrt {21} \right) \left( {\frac {y}{x}}+{\frac {z}{y}}+{\frac {x}{z}}-3 \right)\]

sqing wrote: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=512906#p512906: \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \] just expanding and AM GM

Solution of Iran 2005 $\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\iff\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2\frac{a}{c}+2\frac{b}{a}+2\frac{c}{b}\ge 3+\frac{a+c}{b}+\frac{a+b}{c}+\frac{c+b}{a}\iff \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} $ Which is true because by AM-GM $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\ge 3$ also we have $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ because of Nordic Math Contest 1987 ;D

Let $a,b,c$ be positive numbers such that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{49}{4}$$Find all possible values of the expression $$E=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$$. Problem O438 from Mathematical Reflections Issue 1 - 2018 Proposed by Marius Stănean, Zalău, România

WolfusA wrote: Let $a,b,c$ be positive numbers such that $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{49}{4}$$Find all possible values of the expression $$E=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$$. Problem O438 from Mathematical Reflections Issue 1 - 2018 Proposed by Marius Stănean, Zalău, România I think this problem has been released for a while.(I have seen before) Let $M= \sum_{cyc}\frac{a}{b}$ and $N= \sum_{cyc}\frac{b}{a}$ From the given condition we have: $M+N= \frac{37}{4}$ We want to find the bound of possible values of $E= \sum_{cyc}\frac{a^2}{b^2}=M^2-2N=M^2+2M-\frac{37}{2}$. Consider cubic polynomial $P(t)$ that have 3 positive real roots:$ x=\frac{a}{b} , y=\frac{b}{c} ,z=\frac{c}{a}$ $P(t)=t^3-Mt^2+Nt-1= t^3-Mt^2+(\frac{37}{4}-M)t-1$ Its discriminant = $((-M)(\frac{37}{4}-M))^2+18(-M)(\frac{37}{4}-M)(-1)-4(\frac{37}{4}-M)^3-4(-M)^3(-1)-27(-1)^2 =\frac{1}{16}(4M-17)(M-5)(4M^2-37M-601)$ must be nonnegative. Because $0 \le M \le \frac{37}{4}$ so $4M^2-37M-601=M(4M-37)-601 \le -601 < 0$ and $(4M-17)(M-5) \le 0 , M \in [\frac{17}{4},5]$ We need to check that all $M \in [\frac{17}{4},5]$ cause $P(t)$ to have only positive real roots. Consider $t \le 0$, we have $t^3,-Mt^2,(\frac{37}{4}-M)t \le 0$ and $-1<0$ so $P(t) < 0$ for all $t \le 0$ Hence all roots of $P(t)$ are positive, so $a,b,c>0$ is valid. So $E= M^2+2M-\frac{37}{2} \in [\frac{129}{16},\frac{33}{2}]$ The lowest value $\frac{129}{16}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{4},2,2)$ or permutations. The greatest value $\frac{33}{2}$ is occurred when $(\frac{a}{b},\frac{b}{c},\frac{c}{a})=(\frac{1}{2},\frac{1}{2},4)$ or permutations.

Nice trick with discriminant.

Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] Crux Mathematicorum, Vol. 43(3), March 2017: Let $a, b,c \in R^+$ such that $abc = 1. $Prove that $$a^2b + b^2c + c^2a \geq\sqrt{(a + b + c)(ab + bc + ca)} $$$$\iff\left(\frac ba+\frac cb+\frac ac\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$

Attachments:

Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] $$\left(\frac ab+\frac bc+\frac ca\right)^2= (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)+\left(\frac1{ab}+\frac1{b^2}\right)(a-b)^2+\left(\frac1{bc}+\frac1{c^2}\right)(b-c)^2+\left(\frac1{ca}+\frac1{a^2}\right)(c-a)^2.$$

Note that $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^4$ and $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^2(ab+bc+ca)^2 \ge (a+b+c)^4$. Note that we need to prove $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^4 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2(a+b+c)^2$ and we have $(\frac {a}{b}+\frac {b}{c}+\frac {c}{a})^4(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2(ab+bc+ca)^2 \ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^4(a+b+c)^4$ so we need to prove $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2(a+b+c)^2 \ge (\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})^2(ab+bc+ca)^2$ which is true.

one nice!

My solution from WOOT: Looking at the quantities, we are motivated to expand both sides of the inequality. We have $$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right)^2 \ge (a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$$ $$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + 2\left(\frac{b}{a} + \frac{a}{c} + \frac{c}{b} \right) \ge 3 + \frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b}$$ $$\iff \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{b}{a} + \frac{a}{c} + \frac{c}{b} \ge 3 + \frac{a}{b} + \frac{c}{a} + \frac{b}{c}$$Substituting $x = \frac{a}{b}, y = \frac{b}{c}$ and $z = \frac{c}{a}$, it suffices to prove that $$x^2 + y^2 + z^2 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge 3 + x + y + z$$Now note that $$x^2 + \frac{1}{x} \ge 1 + x$$$$\iff (x - 1)^2(x + 1) \ge 0$$which is true. Adding across $x, y, z$ gives us our earlier result, finishing off the problem. Note that $x, y, z > 0$, with equality occurring when $x = y = z = 1$, or $a = b = c$. Our solution is complete.

Let $a,b,c $ be positive real numbers . Prove that $$\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$

Omid Hatami wrote: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\] Let $a,b,c>0.$ Prove that $$ \left(\frac ab+\frac bc+\frac ca\right)^2\geq \frac{a+2b}{c}+\frac{b+2c}{a}+\frac{c+2a}{b} \geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)$$