Question

If x>0,y>0 & z>0, then prove that (×+y)(y+z)(z+x)$\geqslant$8xyz (JEE-MAINS)​

Answer 1

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{A.M\geqslant\:G.M}}}$

TO SOLVE THIS QUESTION WE KNOW THAT A.M IS GREATER THAN OR EQUALS TO G.M .

SO, FROM THIS WE MAKE THREE EQUATION THEN MULTIPLY THOSE EQN.

$\huge{\boxed{\sf{PROOF-}}}$

$\frac{x + y}{2} \geqslant \sqrt{xy} - - - - - (1) \\ \frac{y + z}{2} \geqslant \sqrt{yz} - - - - - (2) \\ \frac{z + x}{2} \geqslant \sqrt{zx} - - - - - (3) \\ (1) \times (2) \times (3) \: we \: get \\ =)( \frac{x + y}{2}) \times (\frac{y + z}{2}) \times ( \frac{z + x}{2}) \geqslant \sqrt{x \times y \times y \times z \times z \times x} \\=) ( \frac{x + y}{2}) \times (\frac{y + z}{2}) \times ( \frac{z + x}{2}) \geqslant xyz \\ =)(x + y)(y + z)(z + x) \geqslant 8xyz$

$\huge{\boxed{\sf{PROVED}}}$

Answer 2

Answer:

Please find the attached images.

Step-by-step explanation: