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Albert
Well-known member
- Jan 25, 2013
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$x,y,z\in R$
$x+y+z=5---(1)$
$xy+yz+zx=3---(2)$
find $\max(z)$
$x+y+z=5---(1)$
$xy+yz+zx=3---(2)$
find $\max(z)$
thanks , your answer is correctMy solution:
$\bf{x+y+z=5 \Rightarrow x+y = 5-z...............(1)}$
$\bf{xy+yz+xz=3}$
$\bf{(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25 \Rightarrow x^2+y^2+z^2=19}$
So $\bf{x^2+y^2 = 19-z^2...............(2)}$
Using Cauchy - Schwartz Inequality
$\bf{\left(x^2+y^2\right).\left(1^2+1^2\right)\geq \left(x+y \right)^2}$
$\bf{\left(19 - z^2 \right).2 \geq \left( 5-z \right)^2}$
$\bf{3z^2-10z-13 \leq 0}$
$\displaystyle \bf{3.\left(z - \frac{13}{3} \right).\left(z+1\right)\leq 0}$
$\displaystyle \bf{ -1 \leq z \leq \frac{13}{3}}$
So $\displaystyle \bf{Max.(z) = \frac{13}{3}}$