Question

Bài 1: cho a, b, c > 0 và a + b + c = 1. tìm GTLN của \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\)

Bài 2: cho a, b, c > 0 thỏa mãn a + b + c = 3. CMR: \(\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}>=3\)

Answer 1

Bài 1

\(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}=a-\dfrac{a^2}{a+1}+b-\dfrac{b^2}{b+1}+c-\dfrac{c^2}{c+1}\)

\(=1-\left(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\right)\)

Áp dụng bđt Cauchy dạng phân thức \(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{1}{1+3}=\dfrac{1}{4}\)

\(\Rightarrow1-\left(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\right)\le1-\dfrac{1}{4}=\dfrac{3}{4}\)

\(\Rightarrow GTLN=\dfrac{3}{4}\) Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Bài 2

\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\dfrac{a}{b^2+1}+\dfrac{1}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{1}{c^2+1}+\dfrac{c}{a^2+1}+\dfrac{1}{a^2+1}\)

Xét \(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{a^2c}{a^2+1}\)

Xét \(\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}+\dfrac{1}{a^2+1}=1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}+1-\dfrac{a^2}{a^2+1}\)

\(\Rightarrow P=6-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}+\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)

Áp dụng bđt Cauchy cho 2 số thực dương ta có \(b^2+1\ge2b\Rightarrow\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)

\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ac}{2}\)

Theo hệ quả của bđt Cauchy ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow3\ge ab+bc+ac\) \(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ac}{2}\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{3}{2}\)

Áp dụng bđt Cauchy cho 2 số thực dương ta có \(a^2+1\ge2a\Rightarrow\dfrac{a^2}{a^2+1}\le\dfrac{a^2}{2a}=\dfrac{a}{2}\)

\(\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\)

\(\Rightarrow P\ge6-\left(\dfrac{3}{2}+\dfrac{3}{2}\right)=3\left(đpcm\right)\)

Dấu ''='' xảy ra khi \(a=b=c=1\)

Answer 2

Bài 1 : Ta có : \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}=\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}+\dfrac{c^2}{c^2+c}\)

Theo BĐT CÔ - SI dưới dạng engel ta có :

\(\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}+\dfrac{c^2}{c^2+c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\left(a+b+c\right)}=\dfrac{1}{a^2+b^2+c^2+1}\le\dfrac{1}{\dfrac{1}{a+b+c}+1}=\dfrac{1}{\dfrac{1}{3}+1}=\dfrac{4}{3}\)

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