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Ta có: \(\dfrac{ab}{c+1}=\dfrac{ab}{b+c+a+c}\le\dfrac{1}{4}\left(\dfrac{ab}{b+c}+\dfrac{ab}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại:
\(\dfrac{bc}{a+1}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế các BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )
Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
Thôi đang rảnh, giúp bạn bài này luôn vậy!!
Giải:
Ta có:
\(VT=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{a^2}{a+b}\right)=A+B\)
\(A+3=\dfrac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left[\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right]\)
\(\ge\dfrac{1}{2}3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}3\sqrt[3]{\dfrac{1}{a+b}\dfrac{1}{b+c}\dfrac{1}{c+a}}=\dfrac{9}{2}\)
\(\Rightarrow A\ge\dfrac{3}{2}\)
\(1^2=\left(a+b+c\right)^2\le\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\left(a+b+b+c+c+a\right)\)
\(\Leftrightarrow1\le B.2\Leftrightarrow B\ge\dfrac{1}{2}\)
Từ đó ta có: \(VT\ge\dfrac{3}{2}+\dfrac{1}{2}=2=VP\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a\left(a+b+c\right)+b^2}{b+c}+\dfrac{b\left(a+b+c\right)+c^2}{c+a}+\dfrac{c\left(a+b+c\right)+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+ab+ac+b^2}{b+c}+\dfrac{ab+b^2+bc+c^2}{c+a}+\dfrac{ca+bc+c^2+a^2}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+c^2+b\left(c+a\right)}{c+a}+\dfrac{c^2+a^2+c\left(a+b\right)}{a+b}\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}+1\ge2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{b+c}+\dfrac{b^2+c^2}{c+a}+\dfrac{c^2+a^2}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a+b+c\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\)
Áp dụng bất đẳng thức Mincopski
\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{2\left(a+b+c\right)^2}=\sqrt{2}\)
\(\Rightarrow\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2\ge2\)
\(\Rightarrow\dfrac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2}\ge1\)
\(\Rightarrow\dfrac{\sqrt{\left(a^2+b^2\right)^2}}{b+c}+\dfrac{\sqrt{\left(b^2+c^2\right)^2}}{c+a}+\dfrac{\sqrt{\left(c^2+a^2\right)^2}}{a+b}\ge1\)
\(\Leftrightarrow\dfrac{a+b^2}{b+c}+\dfrac{b+c^2}{c+a}+\dfrac{c+a^2}{a+b}\ge2\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{2a+b+c}=\dfrac{bc}{a+b+a+c}\le\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{ca}{a+2b+c}=\dfrac{ca}{a+b+b+c}\le\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{ab}{a+b+2c}=\dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{bc}{4\left(a+b\right)}+\dfrac{bc}{4\left(a+c\right)}+\dfrac{ca}{4\left(a+b\right)}+\dfrac{ca}{4\left(b+c\right)}+\dfrac{ab}{4\left(a+c\right)}+\dfrac{ab}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\left[\dfrac{bc}{4\left(a+b\right)}+\dfrac{ca}{4\left(a+b\right)}\right]+\left[\dfrac{bc}{4\left(a+c\right)}+\dfrac{ab}{4\left(a+c\right)}\right]+\left[\dfrac{ca}{4\left(b+c\right)}+\dfrac{ab}{4\left(b+c\right)}\right]\)
\(\Rightarrow VT\le\dfrac{bc+ca}{4\left(a+b\right)}+\dfrac{bc+ab}{4\left(a+c\right)}+\dfrac{ca+ab}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{c\left(a+b\right)}{4\left(a+b\right)}+\dfrac{b\left(c+a\right)}{4\left(a+c\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{a+b+c}{4}\)
\(\Leftrightarrow\dfrac{bc}{2a+b+c}+\dfrac{ca}{a+2b+c}+\dfrac{ab}{a+b+2c}\le\dfrac{a+b+c}{4}\) ( đpcm )
Ta có: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}+2a+2b+2c\)
\(=\left(\dfrac{a^3}{bc}+b+c\right)+\left(\dfrac{b^3}{ca}+a+c\right)+\left(\dfrac{c^3}{ab}+a+b\right)\ge3\sqrt[3]{\dfrac{a^3}{bc}.b.c}+3\sqrt[3]{\dfrac{b^3}{ca}.a.c}+3\sqrt[3]{\dfrac{c^3}{ab}.a.b}=3a+3b+3c\)
\(\Rightarrow\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}+2a+2b+2c\ge3a+3b+3c\)
\(\Rightarrow\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge a+b+c\)
\("="\Leftrightarrow a=b=c\)
Ta có: \(A=\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)(do \(a;b;c>0\) )
Áp dụng BĐT \(a^2+b^2+c^2\ge ab+bc+ca\)(\("="\Leftrightarrow a=b=c\))
\(A=\dfrac{a^4+b^4+c^4}{abc}=\dfrac{\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2}{abc}\ge\)
\(\ge\dfrac{a^2b^2+b^2c^2+c^2a^2}{abc}\ge\dfrac{abc\left(a+b+c\right)}{abc}=a+b+c\)
1) Áp dụng bất đẳng Bunyakovsky dạng cộng mẫu ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)
\(=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)
(Cauchy 3 số) Dấu "=" xảy ra khi: a = b = c
2) Áp dụng kết quả phần 1 ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^2+c^3\right)^2}{3\cdot\frac{1}{3}}=\left(a^3+b^3+c^3\right)^2\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(VT=\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ca}{c+a}+\dfrac{c\left(a+b+c\right)+ab}{a+b}\)
\(VT=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{c+a}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\)
Ta có:
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{c+a}\ge2\left(a+b\right)\)
Tương tự: \(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+c\right)\)
\(\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(b+c\right)\)
Cộng vế với vế:
\(\Rightarrow VT\ge2\left(a+b+c\right)=2\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
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