Fascinating minima

Algebra Level 4

( x + 1 y ) 2 + ( y + 1 z ) 2 + ( z + 1 x ) 2 \large \left (x+\dfrac{1}{y}\right)^2+\left(y+\dfrac{1}{z}\right)^2+\left(z+\dfrac{1}{x}\right)^2

If x , y x,y and z z are are positive numbers satisfying x + y + z = 6 x+y+z=6 , find the minimum value of expression above.

For more problems try my set .


The answer is 18.75.

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Relevant wiki: Titu's Lemma

By Titu's Lemma ,

( x + 1 y ) 2 + ( y + 1 z ) 2 + ( z + 1 x ) 2 ≥ ( x + y + z + 1 x + 1 y + 1 z ) 2 3 \large \left (x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2 \ge \frac{(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{3} ..... (1)

By AM ≥ HM \text{AM}\ge \text{HM} ,

x + y + z 3 ≥ 3 1 x + 1 y + 1 z ⟹ 1 x + 1 y + 1 z ≥ 9 x + y + z ⟹ 1 x + 1 y + 1 z ≥ 3 2 \frac{x+y+z}{3}\ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\implies \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \ge \frac{9}{x+y+z}\implies \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{3}{2}

Plugging it in 1 we get ,

( x + 1 y ) 2 + ( y + 1 z ) 2 + ( z + 1 x ) 2 ≥ ( 6 + 3 2 ) 2 3 = 18.75 \large \left (x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2 \ge \frac{(6+\frac{3}{2})^2}{3} = \boxed{18.75}

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You can just think of the application of Titu's Lemma as an application of Cauchy-Schwarz inequality on: ( x + 1 y , y + 1 z , z + 1 x ) , ( 1 , 1 , 1 ) \left( x+\frac{1}{y},y+\frac{1}{z},z+\frac{1}{x} \right),\left (1,1,1 \right)

Sam Bealing - 5 years, 1 month ago

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Obviously, the titu's lemma is just an extended version of the CS inequality. So the titu's llemma here gives an one line conclusion here. :)

Aditya Narayan Sharma - 5 years, 1 month ago

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