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\(BĐT\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right)\ge\frac{9}{4}\)
Áp dụng BĐT Bunhi kết hợp với Nesbit :
\(VT=\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{c+a}\right)^2+\left(\frac{\sqrt{c}}{a+b}\right)^2\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
Vậy BĐT đc chứng minh . Dấu bằng xảy ra khi \(a=b=c\)
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{c+a}\right)^2+\left(\frac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dúng bất đẳng thức Bunhiacopxki ta có :
\(VT\ge\left(\sqrt{a}.\frac{\sqrt{a}}{b+c}+\sqrt{b}.\frac{\sqrt{b}}{c+a}+\sqrt{c}.\frac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)
Xét \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Áp dụng bất đẳng thức Cauchy dạng phân thức ta có :
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)
\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(\Rightarrow VT\ge\frac{9}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c\)
Chúc bạn học tốt !!!
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ta có: \(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}.\)
\(\ge3\sqrt[3]{\frac{a.b.c}{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}=\frac{3}{\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}}\) (vì abc=1) (*)
Mặt khác: \(\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2\ge64abc=64=4^3\) (vì abc=1)
=> \(\sqrt[3]{\left(a+1\right)^2.\left(b+1\right)^2.\left(c+1\right)^2}\ge4\) (**)
Từ (*), (**)=> đpcm
Bạn dưới kia làm ngược dấu thì phải,mà bài này hình như là mũ 3
\(\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{b+1}{8}\ge3\sqrt[3]{\frac{a^3\left(a+1\right)\left(b+1\right)}{64\left(a+1\right)\left(b+1\right)}}=\frac{3a}{4}\)
Tương tự rồi cộng lại:
\(RHS+\frac{2\left(a+b+c\right)+6}{8}\ge\frac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow RHS\ge\frac{3}{4}\) tại a=b=c=1
Bất đẳng thức
<=> \(\frac{a\left(a+b+c\right)}{\left(b+c\right)^2}+\frac{b\left(a+b+c\right)}{\left(c+a\right)^2}+\frac{c\left(a+b+c\right)}{\left(a+b\right)^2}\ge\frac{9}{4}\)
VT = \(\left(\frac{a^2}{\left(b+c\right)^2}+\frac{b^2}{\left(a+c\right)^2}+\frac{c^2}{\left(a+b\right)^2}\right)+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\ge\frac{1}{3}.\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)^2+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
lại có:
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)
=> VT\(\ge\frac{1}{3}.\left(\frac{3}{2}\right)^2+\frac{3}{2}=\frac{9}{4}\)
Dấu "=" xảy ra <=> a = b = c.
Hoặc em có thể áp dụng Bunhia
bất đẳng thức
<=> \(\left(a+b+c\right)\left(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right)\ge\frac{9}{4}\)
VT\(\ge\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)