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Do a+b+c= 0
<=> a+b= -c
=> (a+b)2= c2
Tương tự: (c+a)2= b2, (c+b)2= a2
Ta có: \(A=\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}+\frac{1}{a^2+b^2-c^2}\)
\(=\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{c^2+a^2-\left(c+a\right)^2}+\frac{1}{a^2+b^2-\left(a+b\right)^2}\)
\(=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}\)
\(=\frac{a+b+c}{-2abc}=0\)
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a) Áp dụng Cauchy Schwars ta có:
\(M=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{\left(a+b+c\right)^2}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
b) \(N=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\ge\frac{\left(1+2+3\right)^2}{a+b+c+3}=\frac{36}{6}=6\)
Dấu "=" xảy ra khi: x=y=1
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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)
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Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với 0, không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow ab\ge0\)
Mặt khác do \(c\le1\Rightarrow\left\{{}\begin{matrix}1-c^2\ge0\\1-c\ge0\end{matrix}\right.\)
\(\Rightarrow2ab\left(1-c\right)+1-c^2\ge0\)
\(\Leftrightarrow2ab+1\ge2abc+c^2\)
\(\Leftrightarrow a^2b^2+2ab+1\ge a^2b^2+2abc+c^2\)
\(\Leftrightarrow\left(ab+c\right)^2\le\left(1+ab\right)^2\le\left(1+a^2\right)\left(1+b^2\right)\) (1)
Từ giả thiết:
\(a^2+b^2+c^2\le1+2abc\Leftrightarrow a^2b^2-2abc+c^2\le1-a^2-b^2+a^2b^2\)
\(\Leftrightarrow\left(ab-c\right)^2\le\left(1-a^2\right)\left(1-b^2\right)\) (2)
Nhân vế với vế (1) và (2):
\(\left(ab+c\right)^2\left(ab-c\right)^2\le\left(1+a^2\right)\left(1+b^2\right)\left(1-a^2\right)\left(1-b^2\right)\)
\(\Leftrightarrow1+2a^2b^2c^2\ge a^4+b^4+c^4\) (đpcm)
Dấu "=" xảy ra khi 1 số bằng 1 và 2 số bằng nhau
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\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
=>\(2\left(ab+bc+ac\right)=0\)
=>ab+bc+ac=0
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)
=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)
=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)
=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)
=>0=0(đúng)
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\(a+b+c=abc\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}=1\Leftrightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)
Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\\ \Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\\ \Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
bình phương (1/a+1/b+1/c) rồi áp dụng HĐT tính bình thường