Cho a, b, c lớn hơn 2 thoả mãn \(\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=2\)
Tìm GTLN của \(H=\left(a-2\right)\left(b-2\right)\left(c-2\right)\)
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\(\frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=2\)
\(\Leftrightarrow\frac{1}{a-1}=\left(1-\frac{1}{b-1}\right)+\left(1-\frac{1}{c-1}\right)\)
\(\Leftrightarrow\frac{1}{a-1}=\frac{b-2}{b-1}+\frac{c-2}{c-1}\)
Áp dụng BĐT Cauchy ta có : \(\frac{1}{a-1}=\frac{b-2}{b-1}+\frac{c-2}{c-1}\ge2\sqrt{\frac{b-2}{b-1}.\frac{c-2}{c-1}}\)
Tương tự : \(\frac{1}{b-1}\ge2\sqrt{\frac{a-2}{a-1}.\frac{c-2}{c-1}}\)
\(\frac{1}{c-1}\ge2\sqrt{\frac{b-2}{b-1}.\frac{a-2}{a-1}}\)
Nhân các BĐT theo vế :
\(\frac{1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\ge\frac{8\left(a-2\right)\left(b-2\right)\left(c-2\right)}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)
\(\Leftrightarrow8\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\Leftrightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le\frac{1}{8}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{5}{2}\)
Vậy maxH = 1/8 <=> a = b = c = 5/2
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)
Ta có:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{6a-b-c-2}{8}\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}\ge\frac{6b-c-a-2}{8}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6c-a-b-2}{8}\end{cases}}\)
Cộng vế theo vế ta được
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{6a-b-c-2}{8}+\frac{6b-c-a-2}{8}+\frac{6c-a-b-2}{8}\)
\(=\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{2}.\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
Theo bđt Cauchy - Schwart ta có:
\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)
\(=\frac{\left(ab+bc+ca\right)^2}{abc\left(ab+bc+ca\right)+3a^2b^2c^2}\)
Đặt \(ab+bc+ca=x;abc=y\).
Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)
\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )
Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1
\(\frac{1}{a-1}=\left(1-\frac{1}{b-1}\right)+\left(1-\frac{1}{c-1}\right)=\frac{b-2}{b-1}+\frac{c-2}{c-1}\ge2\sqrt{\frac{\left(b-2\right)\left(c-2\right)}{\left(b-1\right)\left(c-1\right)}}\)
Tương tự với \(\frac{1}{b-1};\text{ }\frac{1}{c-1}\)
Rồi nhân theo vế 3 bất đẳng thức:
\(\frac{1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\ge8\sqrt{\frac{\left(a-2\right)^2\left(b-2\right)^2\left(c-2\right)^2}{\left(a-1\right)^2\left(b-1\right)^2\left(c-1\right)^2}}=8\frac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)
\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le\frac{1}{8}\)
Vậy GTLN của H là 0,125.
Đẳng thức xảy ra khi \(a=b=c=\frac{5}{2}.\)