If a- b- c - 0- then prove that a3-b3-c3-1a3b3c3 - 4-

We know that A.M ≥ G.M for any set of values.consider a^3 , b^3 , c^3 , 1a^3b^3c^3 #160;A.M =a^3+b^3+c^3+1a^3b^3c^34 G.M =√(a^3.b^3.c^3. ...

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Integration by Parts

If a, b, c >...

Question

If a, b, c $> 0$ , then prove that $(a^{3} + b^{3} + c^{3} + \frac{1}{a^{3} b^{3} c^{3}}) \geq 4$ .

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a + b + c = 0

a^{3} + b^{3} + c^{3} = 3 a b c

a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} =

a^{3} (c - b) + b^{3} (a - c) + c^{3} (b - a)

\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 {(x + y + z)}^{3}}{{(a + b + c)}^{3}}

a + b + c = 5

a b + b c + c a = 10

a^{3} + b^{3} + c^{3} - 3 a b c = - 25

\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

\frac{x^{3}}{a^{3}} + \frac{y^{3}}{b^{3}} + \frac{z^{3}}{c^{3}} = \frac{3 x y z}{a b c}

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