If A+B/b = 3/2 And C/b

     
How to lớn prove $(a+b+c)^2 > 3(bc+ca+ab)$ when a, b, c are positive, unequal & of course, real.

The only thing I have been able to vì chưng is:

$a^2+b^2+c^2 > 0$

So $(a+b+c)^2 - 2 (ab+bc+ac) >0$

$(a+b+c)^2 > 2(ab+bc+ac) $

I need one more $(ab+bc+ac)$ on the right-hand side, but I don"t see any way to vì it.I tried using A.M. - G.M. Inequality, but even after trying many terms, I failed t obtain it.

I would prefer an answer without Cauchy-Schwarz inequality, using only A.M. - G.M. Inequality if it is required.




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$$(a+b+c)^2 - 3(ab + bc + ac) = a^2+b^2+c^2-ab-bc-ac=\=frac12(a^2 -2ab+b^2) + frac12(b^2 -2bc+c^2) + frac12(a^2 -2ac+c^2)=\frac12left((a-b)^2+(b-c)^2+(a-c)^2 ight) geq 0.$$


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Notice $2 ( a^2 + b^2 + b^2 - ab -bc - ac ) = (a-b)^2 + (a-c)^2 + (b-c)^2 geq 0 $

Hence,

$$ a^2 + b^2 + c^2 geq ab + bc + ac $$

Now, use this và the fact that

$$ (a+b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc+ac) $$




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For$$W= m AB+ m C$$$$W= -, m AD+ m E$$$$ m B,, m C,, m D,, m Egeqq 0, herefore,W= frac m BE+ m CD m B+ m Dgeqq 0 ag29$$Such as$$W= sum,a^,2- sum,ab= left ( a- c ight )^,2+ left ( a- b ight )left ( c- b ight )geqq 0$$$$W= sum,a^,2- sum,ab= left ( a+ c- 2,b ight )^,2+ 3left ( a- b ight )left ( b- c ight )geqq 0$$$$ herefore,W= frac3(,a- c,)^,2+ (,a+ c- 2,b,)^,24geqq 0$$I gọi it $lceil$ DRIVE!S.O.S $ floor$


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