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\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt Cô Si: \(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
Tương tự,cộng theo vế và rút gọn =>đpcm
\(\frac{a+bc}{b+c}+\frac{b+ac}{c+a}+\frac{c+ab}{a+b}\)
\(=\frac{a\left(a+b+c\right)+bc}{b+c}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{c\left(a+b+c\right)+ab}{a+b}\)
\(=\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}+\frac{\left(c+a\right)\left(c+b\right)}{a+b}\)
Áp dụng bđt CÔ si
\(\frac{\left(a+b\right)\left(a+c\right)}{b+c}+\frac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\)
.............
Bài 1:
Ta có: a + b - 2c = 0
⇒ a = 2c − b thay vào a2 + b2 + ab - 3c2 = 0 ta có:
(2c − b)2 + b2 + (2c − b).b − 3c2 = 0
⇔ 4c2 − 4bc + b2 + b2 + 2bc − b2 − 3c2 = 0
⇔ b2 − 2bc + c2 = 0
⇔ (b − c)2 = 0
⇔ b − c = 0
⇔ b = c
⇒ a + c − 2c = 0
⇔ a − c = 0
⇔ a = c
⇒ a = b = c
Vậy a = b = c
\(A=\left(1+b^2+a^2+a^2b^2\right).\left(1+c^2\right)\)
\(=1+a^2+b^2+c^2+a^2c^2+b^2c^2+a^2b^2+a^2b^2c^2\)
\(=1+\left(a+b+c\right)^2-2.\left(ab+bc+ac\right)+\left(ab+bc+ac\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
Thay ab+bc+ac=1 vào A, ta có:
\(A=1+\left(a+b+c\right)^2-2+1-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c\right)^2-2abc.\left(a+b+c\right)+a^2b^2c^2\)
\(=\left(a+b+c-abc\right)^2\)
Vì a,b,c thuộc Z
\(\Rightarrow\left(a+b+c-abc\right)^2\)là số chính phương
\(\hept{\begin{cases}\left(1+a^2\right)=\left(ab+bc+ca+a^2\right)=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(a+c\right)\\\left(1+b^2\right)=\left(ab+bc+ca+b^2\right)=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\\\left(1+c^2\right)=\left(ab+bc+ca+c^2\right)=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\end{cases}}\)
\(\Rightarrow A=\text{[}\left(a+b\right)\left(b+c\right)\left(c+a\right)\text{]}^2\Rightarrow\text{đ}pcm\)
Ta có : \(a^2+b^2\ge ab+1\)
\(2\sqrt{a^2b^2}\ge ab+1\)
\(ab\ge1\)
Dấu = xảy ra \(< =>a=b=\sqrt{1}=1\)
Bđt ngược dấu rồi thì phải
Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)(1)
<=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)(Vì a + b + c = 9)
<=> \(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\)
<=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)
Lại có \(\frac{a}{b}+\frac{b}{a}\ge2\)
<=> \(\frac{a^2+b^2}{ab}\ge2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\left(\text{đúng}\right)\)
Tương tự \(\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\\\frac{b}{c}+\frac{c}{b}\ge2\end{cases}}\)
<=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(đúng)
=> (1) được chứng minh
Áp dụng bđt Svac-xơ ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}=\frac{9}{9}=1\) ( Vì a+b+c=1)