Merry Christmas 2017!

Algebra Level 3

If $a, b, c> 0$ and $a + b + c = 6$ , find the possible value of

$\large\ \sum_{\text{cyc}} { { \left( a + \frac { 1 }{ b } \right) }^{ 2 } }$

10 35 15 12

1 solution

Chew-Seong Cheong
Dec 19, 2017

$\begin{aligned} \sum_{\text{cyc}} \left(a+\frac 1b\right)^2 & \ge \frac {\left(a+b+c+{\color{#3D99F6}\frac 1a + \frac 1b + \frac 1c}\right)^2}3 & \small \color{#3D99F6} \text{Using Titu's lemma again.} \\ & \ge \frac {\left(a+b+c+{\color{#3D99F6}\frac {3^2}{a+b+c}}\right)^2}3 \\ & = \frac {\left(6+{\color{#3D99F6}\frac 96}\right)^2}3 = 18.75 \end{aligned}$

Since $\displaystyle \sum_{\text{cyc}} \left(a+\frac 1b\right)^2 \ge 18.75$ . The only possible value from the options is $\boxed{35}$ .

@Chew-Seong Cheong ,

But sir can't i do this after your 1st step?

$\large\ \frac { { \left( a + b + c + \frac { 1 }{ a } + \frac { 1 }{ b } + \frac { 1 }{ c } \right) }^{ 2 } }{ 3 } \ge \frac { { \left( 6.\sqrt [ 6 ]{ a.b.c.\frac { 1 }{ a } .\frac { 1 }{ b } .\frac { 1 }{ c } } \right) }^{ 2 } }{ 3 } \ge \frac { { 6 }^{ 2 } }{ 3 } = 12$

I have applied $AM \ge GM$ in the first inequality.

Please tell where am i wrong?

Priyanshu Mishra - 3 years, 5 months ago

You are assuming $a=b=c=2$ , then $\dfrac {(6\sqrt{abc(1/a)(1/b)(1/c)})^2}3 = 12$ . But $\dfrac {(a+b+c+1/a+1/b+1/c)^2}3 = \dfrac {(6+3/2)^2}3 \ne 12$ . For all these cases, there must exist $a$ , $b$ and $c$ that equality occurs.

Chew-Seong Cheong - 3 years, 5 months ago

Oh i see. Thanks for clarification.

Priyanshu Mishra - 3 years, 5 months ago

Merry Christmas 2017!

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