a-b-c-gt-0-prove-a-a-2-8bc-b-b-2-8ac-c-c-2-8ab-1-help-please-

Question Number 98570 by HamraboyevFarruxjon last updated on 14/Jun/20

a, b, c > 0 p r o v e :

\frac{a}{\sqrt{a^{2} + 8 b c}} + \frac{b}{\sqrt{b^{2} + 8 a c}} + \frac{c}{\sqrt{c^{2} + 8 a b}} ⩾ 1

h e l p p l e a s e \dots

Commented by MJS last updated on 14/Jun/20

extremes where a = b = c

here this leads to

3 \frac{a}{\sqrt{a^{2} + 8 a^{2}}} = 1 for a > 0

\Rightarrow \min (lhs) = 1 \Rightarrow proven

Answered by 1549442205 last updated on 15/Jun/20

Applying the Cauchi - Schwartz we have

we have P = \frac{a}{\sqrt{a^{2} + 8 bc}} + \frac{b}{\sqrt{b^{2} + 8 ca}} + \frac{c}{\sqrt{c^{2} + 8 ab}} = \frac{a^{2}}{a \sqrt{a^{2} + 8 bc}} + \frac{b^{2}}{b \sqrt{b^{2} + 8 ca}} + \frac{c^{2}}{c \sqrt{c^{2} + 8 ab}} ⩾ \frac{{(a + b + c)}^{2}}{a \sqrt{a^{2} + 8 bc} + b \sqrt{b^{2} + 8 ca} + c \sqrt{c^{2} + 8 ab}}

On the other hands, also by C - S we have

a \sqrt{a^{2} + 8 bc} + b \sqrt{b^{2} + 8 ca} + c \sqrt{c^{2} + 8 ab} = \sqrt{a} . \sqrt{a^{3} + 8 abc} + \sqrt{b} . \sqrt{b^{3} + 8 abc} + \sqrt{c} . \sqrt{c^{3} + 8 abc}

⩽ \sqrt{(a + b + c) (a^{3} + b^{3} + c^{3} + 24 abc)} . Hence,

P ⩾ \frac{{(a + b + c)}^{2}}{\sqrt{(a + b + c) (a^{3} + b^{3} + c^{3} + 24 abc)}} = \sqrt{\frac{{(a + b + c)}^{3}}{a^{3} + b^{3} + c^{3} + 24 abc}} . Therefore, it is enough

to prove that \frac{{(a + b + c)}^{3}}{a^{3} + b^{3} + c^{3} + 24 abc} ⩾ 1 \Leftrightarrow

{(a + b + c)}^{3} ⩾ a^{3} + b^{3} + c^{3} + 24 abc \Leftrightarrow (a + b) (b + c) (c + a) ⩾ 8 abc

This final inequality is always true because

it is followed from the inequlities :

a + b ⩾ 2 \sqrt{ab}, b + c ⩾ 2 \sqrt{bc}, c + a ⩾ 2 \sqrt{ca}

Thus, P ⩾ 1 . The equality occurs if an only if

a = b = c (q . e . d)

Commented by Farruxjano last updated on 15/Jun/20

Sir thanks a lot

Commented by 1549442205 last updated on 25/Jun/20

Thank you! you are wellcome, sir .