Squared Inequality

Algebra Level 3

$a,b,c$ are reals satisfying $ab^2+bc^2+ca^2=3$ . If the statement $\dfrac{a^2+b^2}{c^2}+\dfrac{b^2+c^2}{a^2}+\dfrac{c^2+a^2}{b^2}\ge k$ is always true for some real number $k$ , then find the largest possible value of $k$ .

The answer is 6.000.

4 solutions

Karthik Kannan
Jun 22, 2014

The expression whose minimum value has to be found can be written as:

$\left( \displaystyle\frac{a}{b}-\displaystyle\frac{b}{a}\right)^{2}+\left( \displaystyle\frac{b}{c}-\displaystyle\frac{c}{b}\right)^{2}+\left( \displaystyle\frac{c}{a}-\displaystyle\frac{a}{c}\right)^{2}+6\geq6$

Equality occurs when:

$a^{2}=b^{2}, b^{2}=c^{2}, c^{2}=a^{2}$

Thus the following two cases arise:

$\text{Case 1:}$

$a=b=c=x(\text{say})$

$\therefore 3x^{3}=3$

$\therefore x=a=b=c=1$

$\text{Case 2:}$

$\text{a, b, c all have equal magnitudes but one of them (say a) has opposite sign}$

$\therefore -a=b=c=y(\text{say})$

$\therefore -y^{3}+y^{3}+y^{3}=3$

$\therefore y=-a=b=c=3^{\frac{1}{3}}$

How did you find that first expression which is minimum?s

Hafizh Ahsan Permana - 6 years, 11 months ago

$\begin{aligned} \sum_{cyc}\dfrac{a^2+b^2}{c^2}&= \sum_{cyc}\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}\\ &= \sum_{cyc}\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\\ &= \sum_{cyc}\dfrac{a^2}{b^2}-2\left(\dfrac{a}{b}\right)\left(\dfrac{b}{a}\right)+\dfrac{b^2}{a^2}+2\left(\dfrac{a}{b}\right)\left(\dfrac{b}{a}\right)\\ &= \sum_{cyc} \left(\dfrac{a}{b}-\dfrac{b}{a}\right)^2+2\left(\dfrac{a}{b}\right)\left(\dfrac{b}{a}\right)\\ &= \left(\sum_{cyc} \left(\dfrac{a}{b}-\dfrac{b}{a}\right)^2\right)+6\\ &\ge 6\end{aligned}$

Daniel Liu - 6 years, 11 months ago

Good job on finding both equality cases!

Daniel Liu - 6 years, 11 months ago

Interesting. I didn't realize that there were 2 equality cases. The question could have asked for "For how many ordered triples of real numbers, is the minimum achieved". Most would have answered 1.

Calvin Lin Staff - 6 years, 11 months ago

Yes I did the same,Sir

Mehul Chaturvedi - 6 years, 7 months ago

Exactly what I did, no need to place my solution then. Still, amazing.

Sean Ty - 6 years, 11 months ago

That was simple , but 6 wasn't accepted for that damn decimal thing .

Samiur Rahman Mir - 6 years, 8 months ago

Simply typing 6 should have been accepted.

I checked and you did not submit any answer at all.

Calvin Lin Staff - 6 years, 7 months ago

Good solution you simply converted them into squares

Aayush Patni - 6 years, 4 months ago

Abhishek Sinha
Jun 22, 2014

$\sum_{\text{cyc}}\frac{a^2+b^2}{c^2}=(\sum_{\text{cyc}}a^2)(\sum_{\text{cyc}}\frac{1}{a^2})-3 \geq 6$ , where the last equality follows from application of Cauchy-Schwartz inequality. Now the assignment $a=b=c=1$ satisfies the given constraint and achieves the lower bound $6$ with equality.

Tekumsa Gentelguy
Jan 7, 2015

WE HAVE SUMS OF RECIPROCALS, SO AM-GM IS TRAIGHTFORWARD, a^2/c^2 +c^2/a^2+....

Anand Raj
Jun 23, 2014

Take a=b=c=1.......And Put In 2nd Equation.........lol

@anand that is what i did, but solutions are for the one that don't get it so you have to explain everything lol.

Mardokay Mosazghi - 6 years, 11 months ago

Squared Inequality

The answer is 6.000.

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