High school inequality 5 abc+acb+bca≥a+b+c.

[solakis1]

1)Prove without using AM-GM :\(\displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a+b+c\)...... a,b,c >02) Prove without using contradiction :

\(\displaystyle a\leq b\wedge b\leq a\Longrightarrow a=b\)

 

[Albert1]

1)Prove without using AM-GM :\(\displaystyle \frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a+b+c\)...... a,b,c >02) Prove without using contradiction :

\(\displaystyle a\leq b\wedge b\leq a\Longrightarrow a=b\)

my solution:

Spoiler

(1)$\times abc$
we have to prove :$(ab)^2+(bc)^2+(ca)^2>(abc)\times (a+b+c)=a^2bc+b^2ca+c^2ab$
for $(A-B)^2=A^2-2AB+B^2$
$\therefore A^2+B^2>2AB---(3)$
likewise $B^2+C^2>2BC---(4)$
$C^2+A^2>2CA---(5)$
(3)+(4)+(5) we have $A^2+B^2+C^2>AB+BC+CA$
let $A=ab,B=bc,C=ca$
and the proof is done
(2)The statement A ∧ B is true if A and B are both true; else it is false.(definition of ∧:logical conjunction)
by using squeeze theorem we get:
\(\displaystyle a\leq b\wedge b\leq a\Longrightarrow a=b\)

 

[solakis1]

my solution:

Spoiler

(1)$\times abc$
we have to prove :$(ab)^2+(bc)^2+(ca)^2>(abc)\times (a+b+c)=a^2bc+b^2ca+c^2ab$
for $(A-B)^2=A^2-2AB+B^2$
$\therefore A^2+B^2>2AB---(3)$
likewise $B^2+C^2>2BC---(4)$
$C^2+A^2>2CA---(5)$
(3)+(4)+(5) we have $A^2+B^2+C^2>AB+BC+CA$
let $A=ab,B=bc,C=ca$
and the proof is done
(2)The statement A ∧ B is true if A and B are both true; else it is false.(definition of ∧:logical conjunction)
by using squeeze theorem we get:
\(\displaystyle a\leq b\wedge b\leq a\Longrightarrow a=b\)

Sorry , but what is the squeeze theorem ??

 

[Albert1]

Sorry , but what is the squeeze theorem ??

squeeze theorem:
https://en.wikipedia.org/wiki/Squeeze_theorem

 

[solakis1]

squeeze theorem:
https://en.wikipedia.org/wiki/Squeeze_theorem

I am sorry but i fail to see the connection.

If you would care to elaborate a bit more

 

[Albert1]

I am sorry but i fail to see the connection.

If you would care to elaborate a bit more

Spoiler

for example :
since $cos \,x\leq \dfrac {sin\, x}{x}\leq 1$
so $ \dfrac {sin \,x}{x} $ approaches 1 as x approaches 0