Prove that a2b2-b2c2-c2a2 - abca-b-c-

We know AM, GM inequality a^2b^2+b^2c^2≥ 2b^2ac b^2c^2+c^2a^2≥ 2c^2ab c^2a^2+a^2b^2≥ 2a^2bc Adding all and then dividing by 2, we get a^ ...

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Sum of n Terms

Prove that ...

Question

Prove that $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq a b c (a + b + c)$ .

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a^{4} + b^{4} + c^{4} \geq a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

a + b + c = 0

\frac{a^{4} + b^{4} + c^{4}}{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}

a, b, c

x^{3} + q x + r = 0

b^{2} c^{2}, c^{2} a^{2}, a^{2} b^{2}

If $a + b + c = 0$ then prove

$a^{4} + b^{4} + c^{4} = 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$

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x^{3} - p x^{2} + q x - r = 0

\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}

\frac{1}{b^{2} c^{2}} + \frac{1}{c^{2} a^{2}} + \frac{1}{a^{2} b^{2}}

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