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Ta có:
\(\left(a-b\right)^2\left(b-c\right)^2+\left(b-c\right)^2\left(c-a\right)^2+\left(c-a\right)^2\left(a-b\right)^2\)
\(=\left(a^2+b^2+c^2-ab-bc-ca\right)^2\)
\(\Rightarrow A=\sqrt{\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}+\frac{1}{\left(c-a\right)^2}}\)
\(=\sqrt{\frac{\left(a-b\right)^2\left(b-c\right)^2+\left(b-c\right)^2\left(c-a\right)^2+\left(c-a\right)^2\left(a-b\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}\)
\(=\sqrt{\frac{\left(a^2+b^2+c^2-ab-bc-ca\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}}\)
\(=\frac{\left(a^2+b^2+c^2-ab-bc-ca\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Vì \(a,b,c\in Q\)
\(\Rightarrow A\in Q\)
Đặt \(a-b=x,b-c=y,c-a=z\). \(\Rightarrow x+y+z=a-b+b-c+c-a=0\)
Xét \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=A\)
Khi đó A bằng giá trị tuyệt đối của \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) là số hữu tỉ
Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\) \(\Rightarrow x+y+z=0\)
\(\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2.0}=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}}\)
\(=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(=\left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right|\in Q\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\)
\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\) ( do \(a+b+c=0\) )
\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) ( đpcm )
2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
