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- Feb 14, 2012
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Let $x, y, z, a, b, c$ be positive real numbers for which $z \ge y \ge x$ and $c \ge b \ge a$, $x+y+z=a+b+c$, $xyz=abc$ and $z \ge c$. Prove that $x \ge a$.
My solution:Let $x, y, z, a, b, c$ be positive real numbers for which $z \ge y \ge x$ and $c \ge b \ge a$, $x+y+z=a+b+c$, $xyz=abc$ and $z \ge c$. Prove that $x \ge a$.
Let $x, y, z, a, b, c$ be positive real numbers for which $z \ge y \ge x$ and $c \ge b \ge a$, $x+y+z=a+b+c$, $xyz=abc$ and $z \ge c$. Prove that $x \ge a$.
Hi Albert,can you find another sets of values ?
(a<b<c , x<y<z and c<z )
(a+b+c=x+y+z)
(abc=xyz)
with a,b,c,x,y,z>0