Symmetric? Maybe not

Algebra Level 5

3 x y + z + 4 y x + z + 5 z x + y \large \frac{3x}{y+z}+\frac{4y}{x+z}+\frac{5z}{x+y} If x , y x,y and z z are positive reals, find the minimum value of the expression above.

Submit your answer to 3 decimal places.


The answer is 5.809.

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1 solution

P C
Feb 16, 2016

Call the expression E, we've got E + 12 = 3 x y + z + 3 + 4 y x + z + 4 + 5 z x + y + 5 E+12=\frac{3x}{y+z}+3+\frac{4y}{x+z}+4+\frac{5z}{x+y}+5 E + 12 = 1 2 ( y + z + x + z + x + y ) ( 3 y + z + 4 x + z + 5 x + y ) \Leftrightarrow E+12=\frac{1}{2}(y+z+x+z+x+y)\big(\frac{3}{y+z}+\frac{4}{x+z}+\frac{5}{x+y}\big) Now, applying Cauchy-Schwarz Inequality we get E + 12 ( 3 + 4 + 5 ) 2 2 E+12\geq\frac{(\sqrt{3}+\sqrt{4}+\sqrt{5})^2}{2} E ( 3 + 4 + 5 ) 2 2 12 5.809 \Leftrightarrow E\geq\frac{(\sqrt{3}+\sqrt{4}+\sqrt{5})^2}{2}-12\approx 5.809

Nicely done!

Anik Mandal - 5 years, 3 months ago

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