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Bài 1 : Áp dụng BĐT trong tam giác ta có :
\(\left\{{}\begin{matrix}a< b+c\\b< c+a\\c< a+b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2-\left(b-c\right)^2\le a^2\\b^2-\left(c-a\right)^2\le b^2\\c^2-\left(a-b\right)^2\le c^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(a+b-c\right)\left(a-b+c\right)\le a^2\\\left(b-c+a\right)\left(b+c-a\right)\le b^2\\\left(c-a+b\right)\left(c+a-b\right)\le c^2\end{matrix}\right.\)
Nhân từng vế BĐT ta được :
\(\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\le abc\) ( đpcm )
Bài 2 : Theo BĐT Cô - si ta có :
\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\b+c\ge2\sqrt{bc}\\c+a\ge2\sqrt{ca}\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
\(\Rightarrow\dfrac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge abc\) (1)
Theo câu 1 ta lại có :
\(abc\ge\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\)
\(\Leftrightarrow abc\ge\sqrt{abc\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)}\)
Đặt \(x=a+b+c;y=ab+bc+ac;z=abc\)
Suy ra : \(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(\Leftrightarrow2\left(1+z\right)+\sqrt{2\left(x^2+y^2+z^2-2xz-2y+1\right)}\ge x+y+z+1\)
\(\Leftrightarrow2\left(x^2+y^2+z^2-2xz-2y+1\right)\ge\left(x+y-z-1\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2-2xy-2xz+2x+2yz-2y-2z+1\ge0\)
\(\Leftrightarrow\left(x-y-z+1\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu được chứng minh
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 \(a=b=c=\frac{1}{2}\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(c+a\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\)
Áp dụng BĐT Bun :
\(\frac{c^2}{c^2\left(a+b\right)}+\frac{a^2}{a^2\left(b+c\right)}+\frac{b^2}{b^2\left(a+c\right)}+\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{c^2\left(a+b\right)+a^2\left(b+c\right)+b^2\left(a+c\right)+2abc}=...\)
Dấu ''='' xảy ra khi a = b =c
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}.\)
\(=2\left(1+abc\right)+\sqrt{\left[\left(a+1\right)^2+\left(1-a\right)^2\right]\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]}\)
\(\ge2\left(1+abc\right)+\left(a+1\right)\left(b+c\right)+\left(1-a\right)\left(bc-1\right)\)
\(=\left(1+a\right)\left(1+b\right)\left(1+c\right)\)