Question

Cho 3 số dương a,b,c thỏa mãn a + b + c + abc = 4

Chứng minh rằng

\(P=\dfrac{a}{\sqrt{b+c}}+\dfrac{b}{\sqrt{a+c}}+\dfrac{c}{\sqrt{a+b}}\ge\dfrac{a+b+c}{\sqrt{2}}\)

Answer 1

\(P>=\dfrac{\left(a+b+c\right)^2}{a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}}\)

\(\left(a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}\right)^2< =\left(a+b+c\right)\left(2ab+2cb+2ac\right)\)

=>\(P>=\dfrac{\left(a+b+c\right)\sqrt{a+b+c}}{\sqrt{2\left(ab+ac+cb\right)}}\)

Đặt a+b=S; ab=T(T<=S^2/4)

=>c=(4-S)/(P+1)

=>S<=4

(a+b+c)>=ab+bc+ac

=>\(S+\dfrac{4-S}{P+1}>=P+\dfrac{S\left(4-S\right)}{P+1}\)

=>P(P+1-S)<=(S-2)^2

P+1-S<=0

=>VT<=0<=VVP

P+1-S>0

=>P(P+1-S)<=S^2/4(S^2/4+1-S)

=>P(P+1-S)<=S^2/16*(S-2)^2<=(S-2)^2

=>a+b+c>=ab+ac+bc

=>\(P>=\dfrac{a+b+c}{\sqrt{2}}\)