Inequalities # 2

Algebra Level 3

If x x , y y and z z are positive reals and x + y + z = 6 x + y + z = 6 , then enter the minimum value of the following expression

∑ c y c x , y , z ( x + 1 y ) 2 \large \sum_{\mathrm{cyc}}^{x,y,z} \left( x + \dfrac1y \right)^2 If your answer is of the form A B \dfrac{A}{B} for positive coprime integers, then enter the value of A + B A + B .

Source : RMO training camp (2016)


The answer is 79.

Rishu Jaar by Rishu Jaar , 21, India

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1 solution

( x + 1 y ) 2 + ( y + 1 z ) 2 + ( z + 1 x ) 2 ≥ 1 3 ( x + 1 y + y + 1 z + z + 1 x ) 2 Using Titu’s lemma = 1 3 ( 6 + 1 x + 1 y + 1 z ) 2 Note that x + y + z = 6 ≥ 1 3 ( 6 + 3 2 x + y + z ) 2 Using Titu’s lemma again = 75 4 Equality occurs when x = y = z = 2 \begin{aligned} \left(x+\frac 1y\right)^2 + \left(y+\frac 1z\right)^2 + \left(z+\frac 1x\right)^2 & \ge \frac 13 \left(x+\frac 1y + y+\frac 1z + z+\frac 1x\right)^2 & \small \color{#3D99F6} \text{Using Titu's lemma} \\ & = \frac 13 \left(6+ \frac 1x + \frac 1y + \frac 1z \right)^2 & \small \color{#3D99F6} \text{Note that }x+y+z=6 \\ & \ge \frac 13 \left(6+ \frac {3^2}{x+y+z} \right)^2 & \small \color{#3D99F6} \text{Using Titu's lemma again} \\ & = \frac {75}4 & \small \color{#3D99F6} \text{Equality occurs when }x=y=z=2 \end{aligned}

⟹ A + B = 79 \implies A + B = 79

3 Helpful 1 Interesting 2 Brilliant 0 Confused

nicely done sir!(+1)

Rishu Jaar - 3 years, 7 months ago

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You are welcome.

Chew-Seong Cheong - 3 years, 7 months ago

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