PROVE THAT: Data: $a, b, c, x, y, z > 0$ $a,b,c,x,y,z > 0$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ $1/x+1/y+1/z=1$ ?

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PROVE THAT: Data: $a, b, c, x, y, z > 0$ $a,b,c,x,y,z > 0$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ $1/x+1/y+1/z=1$ ?

We must prove:
$a b c < \frac{a^{x}}{x} + \frac{b^{y}}{z} + \frac{c^{z}}{z}$ $abc<a^x/x+b^y/z+c^z/z$

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Answer 1 · 2017-08-28T09:54:45

See below.

Explanation:

With $x_{k} > 0$ $x_k > 0$ , from $n \sum k = 1 x_{k} \geq {(n \prod k = 1 x_{k})}_{k}^{\frac{1}{n}}$ $sum_(k=1)^n x_k ge (prod_(k=1)^n x_k)^(1/n)$ we can derive

$μ_{1} x_{1} + μ_{2} x_{2} + μ_{3} x_{3} \geq x_{1}^{μ_{1}} x_{2}^{μ_{2}} x_{3}^{μ_{3}}$ $mu_1 x_1+mu_2 x_2+mu_3x_3 ge x_1^(mu_1) x_2^(mu_2) x_3^(mu_3)$

with $μ_{1} + μ_{2} + μ_{3} = 1$ $mu_1+mu_2+mu_3=1$ now choosing

${(x_1=a^x),(x_2=b^y),(x_3=c^z),(mu_1=1/x),(mu_2=1/y),(mu_3=1/z):}$

we get

$a^x/x+b^y/y+c^z/z ge a b c$

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PROVE THAT: Data: $a, b, c, x, y, z > 0$ $a,b,c,x,y,z > 0$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ $1/x+1/y+1/z=1$ ?

We must prove:
$a b c < \frac{a^{x}}{x} + \frac{b^{y}}{z} + \frac{c^{z}}{z}$ $abc<a^x/x+b^y/z+c^z/z$

1 Answer

Explanation:

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PROVE THAT: Data: a,b,c,x,y,z > 0a,b,c,x,y,z>0a,b,c,x,y,z > 0 1/x+1/y+1/z=11x+1y+1z=11/x+1/y+1/z=1?

We must prove: abc<a^x/x+b^y/z+c^z/zabc<axx+byz+czzabc<a^x/x+b^y/z+c^z/z

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Explanation:

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PROVE THAT: Data: $a, b, c, x, y, z > 0$ $a,b,c,x,y,z > 0$ $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ $1/x+1/y+1/z=1$ ?

We must prove:
$a b c < \frac{a^{x}}{x} + \frac{b^{y}}{z} + \frac{c^{z}}{z}$ $abc<a^x/x+b^y/z+c^z/z$