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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:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
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:
\(VT\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
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: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
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 \(a^2+b^2+c^2\ge ab+bc+ca\)(\("="\Leftrightarrow a=b=c\))
\(A=\dfrac{a^4+b^4+c^4}{abc}=\dfrac{\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2}{abc}\ge\)
\(\ge\dfrac{a^2b^2+b^2c^2+c^2a^2}{abc}\ge\dfrac{abc\left(a+b+c\right)}{abc}=a+b+c\)
Lời giải:
Do \(3=ab+bc+ac\) nên ta có:
\(P=\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\)
\(=\frac{a^3}{b^2+ab+bc+ac}+\frac{b^3}{c^2+ab+bc+ac}+\frac{c^3}{a^2+ab+bc+ac}\)
\(=\frac{a^3}{(b+c)(b+a)}+\frac{b^3}{(c+a)(c+b)}+\frac{c^3}{(a+b)(a+c)}\)
Áp dụng BĐT AM-GM:
\(\frac{a^3}{(b+c)(b+a)}+\frac{b+c}{8}+\frac{b+a}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng các BĐT trên vào và rút gọn:
\(\Rightarrow P+\frac{a+b+c}{2}\geq \frac{3}{4}(a+b+c)\)
\(\Rightarrow P\geq \frac{a+b+c}{4}(1)\)
Ta có một hệ quả quen thuộc của BĐT AM-GM đó là:
\((a+b+c)^2\geq 3(ab+bc+ac)\Leftrightarrow (a+b+c)^2\geq 9\)
\(\Rightarrow a+b+c\geq 3(2)\)
Từ \((1); (2)\Rightarrow P\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
Cách 1
Áp dụng bđt Cauchy ta có
\(\frac{a^3}{b}+b+1\ge3a,\frac{b^3}{c}+c+1\ge3b,\frac{c^3}{a}+a+1\ge3a\)
Cộng từng vế 3 bđt trên ta có
\(A=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a+b+c\right)-3\)
Mặt khác (a+b+c)2+3(a+b+c)\(\ge\)18 (biến đổi tương đương là c/m được)
Đặt m=a+b+c
=> t2+3t-18\(\ge\)0
=> t\(\ge\)3
=> A\(\ge\)3
Dấu "=" xảy ra khi a=b=c=1
Cách 2,rất phức tạp :(
\(6=a+b+c+ab+bc+ca\le\frac{\left(a+b+c\right)^2+3\left(a+b+c\right)}{3}\)
Suy ra \(\left(a+b+c\right)^2+3\left(a+b+c\right)-18\ge0\)
\(\Leftrightarrow a+b+c\ge3\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge9\).
Mà \(VT\le3\left(a^2+b^2+c^2\right)\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Leftrightarrow a^2+b^2+c^2\ge3\)
Ta chứng minh BĐT sau = sos cho đẹp: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
\(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{b}-\frac{a^2b}{b}\right)\ge0\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)}{b}-\Sigma_{cyc}a\left(a-b\right)+\Sigma_{cyc}a\left(a-b\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\frac{1}{2}\left(a-b\right)^2\ge0\Leftrightarrow\left(a-b\right)^2\left(\frac{a^2}{b}+\frac{1}{2}\right)\ge0\) (đúng)
Do vậy: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\ge3^{\left(đpcm\right)}\)
Xảy ra đẳng thức khi a = b = c = 1
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 )
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>=3\sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=3\)
\(ab+ac+bc< =3\sqrt{ab\cdot ac\cdot bc}=3abc\)
=>6abc/ab+ac+bc>=2
=>a/b+b/c+c/a+6ab/ab+ac+bc>=5
\(ab+ac+bc\le3\sqrt{ab\cdot ac\cdot bc}\) là sao?