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Áp dụng BĐT AM-GM ta có:
\(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{ab^2+b^2}{b^2+1}\ge\left(a+1\right)-\frac{ab^2+b^2}{2b}=\left(a+1\right)-\frac{ab+b}{2}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\ge a+b+c+3-\frac{a+b+c+ab+bc+ac}{2}\)
\(\ge a+b+c+3-\frac{a+b+c+\frac{\left(a+b+c\right)^2}{3}}{2}\)
\(\ge3+3-\frac{3+\frac{3^2}{3}}{2}=3\)
\("="\Leftrightarrow a=b=c=1\)
Ta có: \(a+1-\frac{a+1}{b^2+1}=\frac{ab^2+b^2}{b^2+1}\le\frac{ab^2+b^2}{2b}=\frac{ab}{2}+\frac{b}{2}\) vì \(b^2+1\ge2b\)
nên \(\frac{a+1}{b^2+1}\ge a+1-\frac{b}{2}-\frac{ab}{2}\) Tương tự:
Vậy ta có: \(VT\ge a+b+c+3-\frac{a+b+c}{2}-\frac{1}{2}\left(ab+bc+ca\right)\)
Vì \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{9}{3}=3\)
nên \(VT\ge3+\frac{a+b+c}{2}-\frac{1}{2}3=3+\frac{3}{2}-\frac{3}{2}=3=VP\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Áp dụng BĐT AM-GM ta có:
\(\frac{a}{1+b^2}=a-\frac{a^2b}{b^2+1}\ge a-\frac{a^2b}{2b}=a-\frac{ab}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{c^2+1}\ge b-\frac{bc}{2};\frac{c}{a^2+1}\ge c-\frac{ca}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
tc \(x^2+y^2\ge2xy\left(cauchy\right)\)
\(\frac{a}{1+b^2}=\frac{a+ab^2-ab^2}{1+b^2}=\frac{a\left(1+b^2\right)-ab}{1+b^2}=a-\frac{ab}{1+b^2}\ge a-\frac{ab}{2ab}\ge a-\frac{1}{2}\)(1)
tương tự \(\frac{b}{1+c^2}\ge b-\frac{1}{2}\)(2)
\(\frac{c}{1+a^2}\ge c-\frac{1}{2}\)(3)
từ (1)(2)(3)=> \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{3}{2}=3-\frac{3}{2}=\frac{3}{2}\left(a+b+c=3\right)\)
=> đpcm
\(\frac{a+1}{b^2+1}=\frac{\left(a+1\right)\left(b^2+1\right)-b^2\left(a+1\right)}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\)
\(\ge a+1-\frac{b^2\left(a+1\right)}{2b}=a+1-\frac{ab+a}{2}\)
Thiết lập các bất đẳng thức tương tự rồi cộng lại ta được:
\(LHS\ge a+b+c+3-\frac{ab+bc+ca+3}{2}\ge6-\frac{\frac{\left(a+b+c\right)^2}{3}+3}{2}=3=RHS\)
Ta có
\(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{ab^2+b^2}{b^2+1}\ge\left(a+1\right)-\frac{ab^2+b^2}{2b}=\left(a+1\right)-\frac{ab+b}{2}\) (1)
Tương tự \(\frac{b+1}{c^2+1}\ge\left(b+1\right)-\frac{bc+c}{2}\) (2)
và \(\frac{c+1}{a^2+1}\ge\left(a+1\right)-\frac{ca+a}{2}\) (3)
Cộng (1), (2), (3) vế theo vế:
\(VT\ge\left(a+b+c+3\right)-\frac{\left(ab+bc+ca\right)+\left(a+b+c\right)}{2}\ge6-\frac{\frac{\left(a+b+c\right)^2}{3}+3}{2}=3\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=1\)