Question

prove that $A.M\geqslant\:G.M\geqslant\:H.M$ [class 11th , chapter -inequalities]​

Answer 1

Step-by-step explanation:

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

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

$A.M-G.M=\frac{a+b}{2}-\sqrt{ab}\\A.M-G.M=)\frac{a+b-2\sqrt{ab}}{2}\\=)A.M-G.M=\frac{{(\sqrt{a}+\sqrt{b}}^{2}}{2}\geqslant\:0$

$\huge{\boxed{\sf{A.M\geqslant\:G.M}}}-----(1)$

$G.M - H.M = \sqrt{ab} - \frac{2ab}{a + b} \\ G.M - H.M = \frac{ \sqrt{ab} }{a + b} \times (a + b - 2 \sqrt{ab} \\ G.M - H.M = (\frac{ \sqrt{ab} }{a + b} ) \times ( \sqrt{a} - \sqrt{b} ) ^{2} \geqslant 0$

$\huge{\boxed{\sf{G.M \geqslant H.M - - - - - (2)}}}$

FROM (1)and (2 )

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

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