question-prove-that-if-a-b-c-R-then-a-b-c-gt-a-b-c-sincerly-yours-M-N-

Question Number 107242 by mnjuly1970 last updated on 09/Aug/20

\dots q u e s t i o n \dots

p r o v e t h a t :

i f a, b, c \in R^{+} t h e n :

♣ \sqrt{a} + \sqrt{b} + \sqrt{c} > \sqrt{a + b + c} ♣

\dots . s i n c e r l y y o u r s \dots

\dots M . N \dots

Answered by abdomathmax last updated on 09/Aug/20

a = x^{2}, b = y^{2}, c = z^{2} (i) \Rightarrow x + y + z > \sqrt{x^{2} + y^{2} + z^{2}}

\Rightarrow {(x + y + z)}^{2} > x^{2} + y^{2} + z^{2} \Rightarrow

x^{2} + y^{2} + z^{2} + 2 (xy + yz + zx) > x^{2} + y^{2} + z^{2}

xy + yz + zx > 0 condition 5 erified if a > 0, b > 0

and c > 0

Commented by mnjuly1970 last updated on 09/Aug/20

t h a n k y o u

Commented by mathmax by abdo last updated on 09/Aug/20

you are welcome

Answered by 1549442205PVT last updated on 09/Aug/20

The inequality \sqrt{a} + \sqrt{b} + \sqrt{c} > \sqrt{a + b + c}

is always true \forall a, b, c \in R^{+} since

squaring two sides we get an equivalent inequality

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

which is always true

Commented by mnjuly1970 last updated on 09/Aug/20

* t h a n k y o u s o m u c h *

t h a n k s

Answered by udaythool last updated on 09/Aug/20

By contradiction method :

If for some a, b, c \in R^{+}

\sqrt{a} + \sqrt{b} + \sqrt{c} ⩽ \sqrt{a + b + c}

\Rightarrow 0 < \sqrt{a b} + \sqrt{a c} + \sqrt{b c} ⩽ 0

which is absurd .

Commented by mnjuly1970 last updated on 09/Aug/20

n i c e v e r y n i c e

Answered by mnjuly1970 last updated on 09/Aug/20

a n w e r : 0 < \frac{a}{a + b + c} < 1 \Rightarrow \frac{a}{a + b + c} < \sqrt{\frac{a}{a + b + c}} *

: 0 < \frac{b}{a + b + c} < \sqrt{\frac{b}{a + b + c}} * *

0 < \frac{c}{a + b + c} < \sqrt{\frac{c}{a + b + c}} * * *

∴ (*) + (* *) + (* * *) \Rightarrow 1 < \frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{\sqrt{a + b + c}}

\sqrt{a + b + c} < \sqrt{a} + \sqrt{b} + \sqrt{c}

q . e . d