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![](https://rs.olm.vn/images/avt/0.png?1311)
Trước hết, với \(a+b+c=1\) ta có:
\(a^2+b^2+c^2=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\)
\(=\left(a^3+ab^2\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+a^2b+b^2c+c^2a\)
\(\ge2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a\)
Hay \(a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Từ đó:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\)
\(\ge\dfrac{3\left(a^2b+b^2c+c^2a\right)\left(a^2+b^2+c^2\right)}{a^2b+b^2c+c^2a}=3\left(a^2+b^2+c^2\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: a+b+c=0
nên a+b=-c
Ta có: \(a^2-b^2-c^2\)
\(=a^2-\left(b^2+c^2\right)\)
\(=a^2-\left[\left(b+c\right)^2-2bc\right]\)
\(=a^2-\left(b+c\right)^2+2bc\)
\(=\left(a-b-c\right)\left(a+b+c\right)+2bc\)
\(=2bc\)
Ta có: \(b^2-c^2-a^2\)
\(=b^2-\left(c^2+a^2\right)\)
\(=b^2-\left[\left(c+a\right)^2-2ca\right]\)
\(=b^2-\left(c+a\right)^2+2ca\)
\(=\left(b-c-a\right)\left(b+c+a\right)+2ca\)
\(=2ac\)
Ta có: \(c^2-a^2-b^2\)
\(=c^2-\left(a^2+b^2\right)\)
\(=c^2-\left[\left(a+b\right)^2-2ab\right]\)
\(=c^2-\left(a+b\right)^2+2ab\)
\(=\left(c-a-b\right)\left(c+a+b\right)+2ab\)
\(=2ab\)
Ta có: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(=\dfrac{a^3+b^3+c^3}{2abc}\)
Ta có: \(a^3+b^3+c^3\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-cb+c^2\right)-3ab\left(a+b\right)\)
\(=-3ab\left(a+b\right)\)
Thay \(a^3+b^3+c^3=-3ab\left(a+b\right)\) vào biểu thức \(=\dfrac{a^3+b^3+c^3}{2abc}\), ta được:
\(M=\dfrac{-3ab\left(a+b\right)}{2abc}=\dfrac{-3\left(a+b\right)}{2c}\)
\(=\dfrac{-3\cdot\left(-c\right)}{2c}=\dfrac{3c}{2c}=\dfrac{3}{2}\)
Vậy: \(M=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)
\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)
\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\)
\(=a\left(b^2c^2-b^2-c^2+1\right)+b\left(a^2c^2-a^2-c^2+1\right)\)
\(+c\left(a^2b^2-a^2-b^2+1\right)\)
\(=ab^2c^2-ab^2-ac^2+a+ba^2c^2-a^2b-bc^2+b\)
\(+ca^2b^2-a^2c-b^2c+c\)
\(=\left(ab^2c^2+ba^2c^2+ca^2b^2\right)+\left(a+b+c\right)\)
\(-\left(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c\right)\)
\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)\)\(-\left[ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\right]\)
\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)+3abc\)\(-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)
\(=abc\left(bc+ac+ab\right)+\left(a+b+c\right)+3abc\)\(-\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(=abc\left(bc+ac+ab\right)+abc+3abc\)\(-abc\left(ab+bc+ca\right)=4abc\)
Vậy \(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cách 3: (rất gọn gàng)
Giả sử \(c=min\left\{a,b,c\right\}\).Trước hết chứng minh: \(4P\le\left(a+b+c\right)^3-3abc\)
Có: \(VP-VT=c\left(\Sigma_{cyc}a^2-\Sigma_{cyc}ab\right)+\left(a-b\right)^2\left(a+b-2c\right)\ge0\)
Vì vậy: \(4P\le\left(a+b+c\right)^3-3abc\le\left(a+b+c\right)^3=1\Rightarrow P\le\frac{1}{4}\)
Đẳng thức xảy ra khi \(\left(a;b;c\right)=\left(\frac{1}{2};\frac{1}{2};0\right)\) và các hoán vị.
P/s: Làm thử, ko chắc, em cũng chưa kiểm tra lại lời giải đâu.
Từ đề bài có \(P=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)=f\left(a;b;c\right)\)
Xét hiệu:
\(f\left(a;b;c\right)-f\left(t;t;c\right)=ab\left(a+b\right)-t^2.\left(2t\right)+bc\left(b+c\right)+ca\left(c+a\right)-2tc\left(t+c\right)\) với \(t=\frac{a+b}{2}\)
Lại có \(b\left(b+c\right)+a\left(c+a\right)-2t\left(t+c\right)\)
\(=b^2+bc+a^2+ca-\left(a+b\right)\left(\frac{a+b}{2}+c\right)\)
\(=\frac{\left(a-b\right)^2}{2}\) nên :
\(f\left(a;b;c\right)-f\left(t;t;c\right)=\frac{c\left(a-b\right)^2}{2}-\left(t^2-ab\right)\left(a+b\right)\)
\(=\frac{2c\left(a-b\right)^2}{4}-\frac{\left(a+b\right)\left(a-b\right)^2}{4}\)
\(=\frac{\left(a-b\right)^2}{4}\left(c-a+c-b\right)\). Không mất tính tổng quát, giả sử \(c=min\left\{a,b,c\right\}\).
Có ngay \(f\left(a;b;c\right)-f\left(t;t;c\right)\le0\) hay \(f\left(a;b;c\right)\le f\left(t;t;c\right)\).
Do đó ta sẽ tìm max của f(t;t;c) = \(2t^3+2tc\left(t+c\right)\). Mặt khác từ đề bài suy ra \(c=1-2t\) mà c> 0 và t > 0do đó \(0\le t\le\frac{1}{2}\)
Do đo \(f\left(t;t;c\right)=2t^3+2t\left(1-2t\right)\left(1-t\right)=6t^3-6t^2+2t\)
Bây giờ xét hiệu \(f\left(t;t;c\right)-\frac{1}{4}=\left(t-\frac{1}{2}\right)\left(6t^2-3t+\frac{1}{2}\right)\le0\forall\)\(0\le t\le\frac{1}{2}\)
Do đó \(f\left(t;t;c\right)\le\frac{1}{4}\).Đẳng thức xảy ra khi \(t=\frac{1}{2}\Rightarrow a=b=\frac{1}{2}\Rightarrow c=0\)
Vậy....
P/s: Em ko chắc vì hoàn toàn chưa kiểm tra lại.