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

AI Thread Summary
The discussion focuses on proving two mathematical inequalities without using specific methods like AM-GM or contradiction. The first inequality states that the sum of the products of variables divided by another variable is greater than or equal to the sum of the variables themselves. The second statement asserts that if two variables are each less than or equal to the other, they must be equal. Participants express confusion about the relevance of the squeeze theorem to these proofs, seeking clarification on its application. The conversation highlights the challenges in understanding and proving these mathematical concepts.
solakis1
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1)Prove without using AM-GM :$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a+b+c$$...... a,b,c >02) Prove without using contradiction :

$$a\leq b\wedge b\leq a\Longrightarrow a=b$$
 
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solakis said:
1)Prove without using AM-GM :$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a+b+c$$...... a,b,c >02) Prove without using contradiction :

$$a\leq b\wedge b\leq a\Longrightarrow a=b$$
my solution:
(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:
$$a\leq b\wedge b\leq a\Longrightarrow a=b$$
 
Last edited:
Albert said:
my solution:
(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:
$$a\leq b\wedge b\leq a\Longrightarrow a=b$$

Sorry , but what is the squeeze theorem ??
 
solakis said:
I am sorry but i fail to see the connection.

If you would care to elaborate a bit more
for example :
since $cos \,x\leq \dfrac {sin\, x}{x}\leq 1$
so $ \dfrac {sin \,x}{x} $ approaches 1 as x approaches 0
 
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