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Sửa đề: Cho a, b, c là các số thực dương thỏa mãn điều kiện abc=1. Chứng minh rằng
\(\frac{1}{ab+b+2}+\frac{1}{bc+c+2}+\frac{1}{ca+a+2}\le\frac{3}{4}\)
Áp dụng bđt Cauchy-Schwarz ta có:
\(\frac{1}{ab+b+2}=\frac{1}{ab+1+b+1}\le\frac{1}{4}\left(\frac{1}{ab+1}+\frac{1}{b+1}\right)\) \(=\frac{1}{4}\left(\frac{abc}{ab\left(1+c\right)}+\frac{1}{b+1}\right)=\frac{1}{4}\left(\frac{c}{1+c}+\frac{1}{b+1}\right)\)
Tương tự \(\frac{1}{bc+c+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{c+1}\right)\)
\(\frac{1}{ca+a+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{a+1}\right)\)
Cộng từng vế các bđt trên ta được
\(VT\le\frac{1}{4}\left(\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\right)=\frac{3}{4}\)
Vậy bđt được chứng minh
Dấu "=" xảy ra khi a=b=c=1
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 \(a=b=c=1\)
\(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự:
\(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Cộng lại:
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{ab}{2}-\frac{bc}{2}-\frac{ca}{2}\)
\(\Rightarrow VT\ge a+b+c\)
Mặt khác:
\(\frac{9}{a+b+c}\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\Rightarrow9\le3\left(a+b+c\right)\Rightarrow a+b+c\ge3\)
Khi đó:
\(VT\ge a+b+c\ge3\left(đpcm\right)\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Áp dụng BĐT Cosi ta có \(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\ge2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)
Tương tự \(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc}\ge1\) \(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ca}\ge1\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được
\(\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\left(\frac{a^2+b^2}{4ab}+\frac{b^2+c^2}{4bc}+\frac{c^2+a^2}{4ca}\right)\ge\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{4b}+\frac{b}{4a}+\frac{b}{4c}+\frac{c}{4b}+\frac{a}{4c}+\frac{c}{4a}\right)\right)\ge\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+c}{b}-\frac{b+c}{a}-\frac{c+a}{b}\right)\ge\frac{3}{4}\)(do \(a+b+c=1\))
\(\Leftrightarrow\frac{3}{4}\ge\frac{3}{4}\) luôn đúng. Từ đó suy ba BĐT được chứng minh. Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
CM BĐT : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\) ( * )
\(\frac{a}{ab+1}=\frac{a\left(ab+1\right)-a^2b}{ab+1}=a-\frac{a^2b}{ab+1}\)
TT ....
Áp dụng BĐT ( * ) với x = \(\sqrt{a}\); y = \(\sqrt{b}\); z = \(\sqrt{c}\) vào bài toán, ta có :
\(\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}=a+b+c-\frac{a^2b}{ab+1}-\frac{b^2c}{bc+1}-\frac{c^2a}{ac+1}\)
\(\ge3-\frac{a^2b}{2\sqrt{ab}}-\frac{b^2c}{2\sqrt{bc}}-\frac{c^2a}{2\sqrt{ac}}=3-\frac{\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
đặt \(\sqrt{\frac{ab}{c}}=x;\sqrt{\frac{bc}{a}}=y;\sqrt{\frac{ca}{b}}=z\Rightarrow xy+yz+zx=1\)
\(P=\frac{ab}{ab+c}+\frac{bc}{bc+a}+\frac{ca}{ca+b}\)
\(=\frac{\frac{ab}{c}}{\frac{ab}{c}+1}+\frac{\frac{bc}{a}}{\frac{bc}{a}+1}+\frac{\frac{ca}{b}}{\frac{ca}{b}+1}=\frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}=\frac{3}{4}\left(Q.E.D\right)\)