Question

1. If a, b, c be positive real numbers, prove that
(i) a^2 +b^2+64 > abc(a + b + c),​

Answer 1

Answer:

$\huge\colorbox{yellow}{Given\:}$

If a, b, c be positive real numbers

$\huge\colorbox{yellow}{To\:Find\:-}$

The minimum value of

$( \frac{1}{a} + \frac{2}{b} + \frac{3}{c})$

$\huge\colorbox{yellow}{Solution\:-}$

$\small\underbrace\mathrm\red{Using\:Formula}$

A. M > GM

Now,

$( \frac{1}{a} + \frac{2}{b} + \frac{3}{c}) \geqslant ( \frac{1}{a} \frac{1}{ {b}^{2} } \frac{1}{ {c}^{3} })^{6} = \frac{1}{2}$

$= > ( \frac{1}{a} + \frac{2}{b} + \frac{3}{c}) \geqslant 3$

Hence, The minimum value is 3

$\huge\colorbox{yellow}{Thank\:You}$

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