tìm gtnn A= ab^2/(a+b)+bc^2/(b+c)+ca^2/(c+a) với a,b,c>0 thỏa mãn a+b+c=3abc
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Ta có: a3+b3+c3=3abc
<=> (a+b+c)(a2+b2+c2-ab-bc-ca)=0
<=> (a+b+c)(2a2+2b2+2c2-2ab-2bc-2ca)=0
<=> (a+b+c)[(a-b)2+(b-c)2+(c-a)2 ] = 0
<=> \(\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
Vì a,b,c phân biệt nên a+b+c=0 => \(\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(c+a\right)\\c=-\left(a+b\right)\end{cases}}\)(*)
Lại có: \(M=\frac{ab^2}{a^2+b^2-c^2}+\frac{bc^2}{b^2+c^2-a^2}+\frac{ca^2}{c^2+a^2-b^2}\)
Thay (*) vào M ta được:
\(M=\frac{-\left(b+c\right)b^2}{\left(b+c\right)^2+\left(b+c\right)\left(b-c\right)}+\frac{-\left(c+a\right)c^2}{\left(c+a\right)^2+\left(c+a\right)\left(c-a\right)}+\frac{-\left(a+b\right)a^2}{\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)}\)
\(=\frac{-\left(b+c\right)b^2}{\left(b+c\right)\left(b+c+b-c\right)}+\frac{-\left(c+a\right)c^2}{\left(c+a\right)\left(c+a+c-a\right)}+\frac{-\left(a+b\right)a^2}{\left(a+b\right)\left(a+b+a-b\right)}\)
\(=\frac{-\left(b+c\right)b^2}{2b\left(b+c\right)}+\frac{-\left(c+a\right)c^2}{2c\left(c+a\right)}+\frac{-\left(a+b\right)a^2}{2a\left(a+b\right)}\)
\(=\frac{-b}{2}-\frac{c}{2}-\frac{a}{2}=\frac{-\left(b+c+a\right)}{2}\)
Mà a+b+c=0
=> M=0
Vậy M=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm GTNN a: $F= 14(a^2+b^2+c^2) + \dfrac{ab+bc+ca}{a^2b+b^2c+c^2a}$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam
Ta có:
\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2b+b^2c+c^2a\right)\le\frac{\left(a^2+b^2+c^2\right)^3}{3}\le\left(a^2+b^2+c^2\right)^4\)
\(\Rightarrow a^2b+b^2c+c^2a\le\left(a^2+b^2+c^2\right)^2\)
Ta lại có:
\(ab+bc+ca=\frac{1-\left(a^2+b^2+c^2\right)^2}{2}\)
Làm tiếp.
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức AM - GM ta có:
\(\dfrac{\sqrt{a^2+b^2+c^2}}{8}+\dfrac{\sqrt{a^2+b^2+c^2}}{8}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{3}{4}\). (1)
Đặt \(\sqrt{a^2+b^2+c^2}=t\Rightarrow\sqrt{\dfrac{4}{3}}\le t\le2\).
\(\dfrac{3\sqrt{a^2+b^2+c^2}}{4}+\dfrac{ab+bc+ca}{2}=\dfrac{3t}{4}+\dfrac{4-2t^2}{4}=\dfrac{\left(2-t\right)\left(2t+1\right)}{4}+\dfrac{3}{2}\ge\dfrac{3}{2}\). (2)
Cộng vế với vế của (1), (2) ta được \(P\ge\dfrac{9}{4}\).
...
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Ta có: \(\sqrt{a^2+b^2+c^2}\ge\sqrt{\dfrac{\left(a+b+c\right)^2}{3}}=\sqrt{3};\sqrt{a^2+b^2+c^2}\le\sqrt{\left(a+b+c\right)^2}=3\).
Đặt \(\sqrt{a^2+b^2+c^2}=t\) \((\sqrt{3}\leq t\leq 3)\).
Ta có: \(P=t+\dfrac{9-t^2}{4}+\dfrac{1}{t^2}=\dfrac{4t^3+9t^2-t^4+4}{4t^2}\).
\(\Rightarrow P-\dfrac{28}{9}=\dfrac{\left(3-t\right)\left(9t^3-9t^2+4t+12\right)}{36}\).
Do \(\sqrt{3}\le t\le3\) nên \(3-t\geq 0\); \(9t^3-9t^2+4t+12>4t+12>0\).
Nên \(P\ge\dfrac{28}{9}\).
Đẳng thức xảy ra khi t = 3, tức (a, b, c) = (0; 0; 3) và các hoán vị.
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x;y;z>0\\x+y+z=3\end{matrix}\right.\)
\(P=\dfrac{x}{\left(3-x\right)^2}+\dfrac{y}{\left(3-y\right)^2}+\dfrac{z}{\left(3-z\right)^2}\)
Ta có đánh giá sau: \(\dfrac{t}{\left(3-t\right)^2}\ge\dfrac{2t-1}{4};\forall t\in\left(0;3\right)\)
Thực vậy, BĐT đã cho tương đương:
\(4t\ge\left(2t-1\right)\left(3-t\right)^2\)
\(\Leftrightarrow-2t^3+13t^2-20t+9\ge0\)
\(\Leftrightarrow\left(9-2t\right)\left(t-1\right)^2\ge0\) (luôn đúng với \(t< 3\))
Áp dụng ta được:
\(P\ge\dfrac{2x-1}{4}+\dfrac{2y-1}{4}+\dfrac{2z-1}{4}=\dfrac{2\left(x+y+z\right)-3}{4}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
Cách khác:
Sau khi đặt ẩn phụ, ta có:
\(P=\dfrac{x}{\left(3-x\right)^2}+\dfrac{y}{\left(3-y\right)^2}+\dfrac{z}{\left(3-z\right)^2}=\dfrac{x}{\left(y+z\right)^2}+\dfrac{y}{\left(z+x\right)^2}+\dfrac{z}{\left(x+y\right)^2}\)
\(\Rightarrow3P=\left(x+y+z\right)\left(\dfrac{x}{\left(y+z\right)^2}+\dfrac{y}{\left(z+x\right)^2}+\dfrac{z}{\left(x+y\right)^2}\right)\ge\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)^2\ge\dfrac{9}{4}\)
(BĐT Netsbitt)
\(\Rightarrow P\ge\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mình quỳ bạn luôn Nhân Thiên Hoàng ạ kiệt lên mạng hỏi mà mày lại bảo vậy thì thua luôn
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Ta có :
\(\frac{a^2}{a+b}=\frac{a\left(a+b\right)-ab}{a+b}=a-\frac{ab}{a+b}\text{≥}a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)(1)
Tương tự : \(\hept{\begin{cases}\frac{b^2}{b+c}\text{≥}b-\frac{\sqrt{bc}}{2}\left(2\right)\\\frac{c^2}{c+a}\text{≥}c-\frac{\sqrt{ac}}{2}\left(3\right)\end{cases}}\)
Cộng vế với vế của (1);(2)(;(3) lại ta được :
\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\text{≥}a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ac}}{2}\)
\(\Leftrightarrow A\text{≥}\left(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ab}\right)+\left(\frac{\sqrt{ab}}{2}+\frac{\sqrt{bc}}{2}+\frac{\sqrt{ac}}{2}\right)\)
Lại lại có : \(a+b+c\text{≥}\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\) (tự chứng minh)
\(\Rightarrow a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ab}\text{≥}0\)
Nên \(A\text{≥}\frac{1}{2}\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=\frac{1}{2}\)có GTNN là 1/2
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab+bc+ca=3abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(Q=\frac{a^2+c^2-c^2}{a\left(c^2+a^2\right)}+\frac{b^2+a^2-a^2}{a\left(a^2+b^2\right)}+\frac{c^2+b^2-b^2}{b\left(b^2+c^2\right)}\)
\(Q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\right)\)
\(Q\ge3-\left(\frac{a}{2ab}+\frac{b}{2bc}+\frac{c}{2ca}\right)=3-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
\(Q_{min}=\frac{3}{2}\) khi \(a=b=c=1\)