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Sửa đề: 1+a^2;1+b^2;1+c^2
\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{a^2+ab+c+ac}}=\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}< =\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
\(\dfrac{b}{\sqrt{1+b^2}}< =\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{b+a}\right)\)
\(\dfrac{c}{\sqrt{1+c^2}}< =\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{a+b}\right)\)
=>\(A< =\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}\)
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT=A+B và xét
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\text{∑}\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\text{∑}\left(3a-\frac{3ab}{2}\right)\)
\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\text{∑}\left(1-\frac{b^2}{1+b^2}\right)\ge\text{∑}\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\text{∑}ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)
(Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))
Dấu = khi a=b=c=1
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT = A + b và xét :
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\Sigma\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\Sigma\left(3a-\frac{3ab}{2}\right)\)\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\Sigma\left(1-\frac{b^2}{1+b^2}\right)\ge\Sigma\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\Sigma ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)( Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)=3}\))
Dấu = khi a = b = c = 1 .
Gọi \(S=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+ab+c^2}+\frac{a^3}{c^2+ab+a^2}\)
Dễ thấy \(P-S=0\)
\(\Rightarrow2P=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+ab+c^2}+\frac{c^3+a^3}{c^2+ab+a^2}\)
Ta chứng minh:
\(\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{a+b}{3}\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng)
\(\Rightarrow2P\ge\frac{a+b}{3}+\frac{b+c}{3}+\frac{c+a}{3}=\frac{2\left(a+b+c\right)}{3}=2\)
\(\Rightarrow P\ge1\)
\(A=\frac{9}{6ab}+\frac{9}{3\left(a^2+b^2\right)}+\frac{1}{2ab}\)
\(\ge\frac{\left(3+3\right)^2}{3\left(a+b\right)^2}+\frac{1}{2\cdot\frac{\left(a+b\right)^2}{4}}\)
\(=\frac{\left(3+3\right)^2}{3\cdot1^2}+\frac{1}{2\cdot\frac{1^2}{4}}=14\)
\("="\Leftrightarrow a=b=\frac{1}{2}\)
Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)
Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)
Cộng vế:
\(P\ge\dfrac{a+b+c}{3}=673\)
Dấu "=" xảy ra khi \(a=b=c=673\)
ta có:
\(c+ab=c.1+ab=c\left(a+b+c\right)+ab=ca+cb+c^2+ab=\left(c+a\right)\left(c+b\right)\)
tương tự như vậy thì \(P=\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)
áp dụng bđt cô si ta có:
\(\frac{a}{a+c}+\frac{b}{b+c}\ge2\sqrt{\frac{ab}{\left(c+a\right)\left(b+c\right)}};\frac{b}{a+b}+\frac{c}{c+a}\ge2\sqrt{\frac{bc}{\left(a+b\right)\left(c+a\right)}};\frac{a}{a+b}+\frac{c}{b+c}\ge2\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+a}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)=\frac{3}{2}\left(Q.E.D\right)\)
Ta có : \(3=ab+bc+ac\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow1\ge abc\)
\(\frac{bc}{a^2\left(b+2c\right)}+\frac{ac}{b^2\left(c+2a\right)}+\frac{ab}{c^2\left(a+2b\right)}\)
\(=\frac{\left(bc\right)^2}{abc\left(ab+2ac\right)}+\frac{\left(ac\right)^2}{abc\left(bc+2ab\right)}+\frac{\left(ab\right)^2}{abc\left(ca+2cb\right)}\)
\(\ge\frac{\left(ab+bc+ac\right)^2}{abc\left(3ab+3ac+3bc\right)}\)\(=\frac{3^2}{9abc}\)\(\ge1\)\(\left(dpcm\right)\)
cm sao bạn
=<3/4