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Dễ chứng minh được \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)\(\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(true\right)\)
\(\Rightarrow2\left(a+b+c\right)\ge\frac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow a+b+c\le6\)
Ta có : \(T=\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\)
\(=1-\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}\)
\(=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(\le3-\frac{9}{a+b+c+3}\le3-\frac{9}{6+3}=2\)
Dấu "=" xảy ra khi \(a=b=c=2\)
2) M = (x25 + 1 + 1 + 1 + 1) - 5x5 + 2
Áp dụng BĐT Cô - si cho 5 số dương x25; 1;1;1;1 ta có: x25 + 1 + 1 + 1 + 1 \(\ge\)5.\(\sqrt[5]{x^{25}.1.1.1.1}=x^5\) = 5x5
=> M \(\ge\) 5x5 - 5x5 + 2 = 2
Vậy M nhỏ nhất = 2 khi x25 = 1 => x = 1
\(ab=\frac{1}{c};c=\frac{1}{ab}\)
\(a+b+c-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+\frac{1}{ab}-\frac{1}{a}-\frac{1}{b}-ab\)
\(=\left(a+b-ab-1\right)+\left(\frac{1}{ab}-\frac{1}{a}-\frac{1}{b}+1\right)\)
\(=-\left(a-1\right)\left(b-1\right)+\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\)
\(=-\left(a-1\right)\left(b-1\right)+\frac{\left(a-1\right)\left(b-1\right)}{ab}\)
\(=-\left(a-1\right)\left(b-1\right)+\left(a-1\right)\left(b-1\right)c\)
\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
Do biểu thức ban đầu dương nên ta có đpcm
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{bc+ac+ab}{abc}=0\Rightarrow bc+ac+ab=0\)
Biến đổi vế phải ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(=a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2+2.0=a^2+b^2+c^2\)
=> ĐPCM
B, -x^2 + 2x - 4 = - ( x^2 - 2x + 4 ) = - ( x^2 - 2x + 1 + 3 ) = -(x + 1 )^2 - 3 <= -3
=> 3/ -(x+1)^2-3 >= 3/-3=-1
Vậy GTNN của A là -1 khi x = -1
Áp dụng BĐT AM-GM ta có:
\(T=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)
\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)
\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)
\(\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}+2\sqrt{a\cdot\frac{1}{2a}}+2\sqrt{b\cdot\frac{1}{2b}}+2\sqrt{\frac{1}{2a}\cdot\frac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu "=" khi \(a=b=\frac{1}{\sqrt{2}}\)
\(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Rightarrow\left(a+b\right)^2\le2\Rightarrow a+b\le\sqrt{2}\Rightarrow\frac{1}{a+b}\ge\frac{\sqrt{2}}{2}\)
\(T=2+a+b+\frac{1}{a}+\frac{1}{b}+\frac{a}{b}+\frac{b}{a}\ge2+a+b+\frac{4}{a+b}+\frac{a}{b}+\frac{b}{a}\)
\(T\ge2+a+b+\frac{2}{a+b}+\frac{a}{b}+\frac{b}{a}+\frac{2}{a+b}\)
\(T\ge2+2\sqrt{\frac{2\left(a+b\right)}{a+b}}+2\sqrt{\frac{ab}{ab}}+2.\frac{\sqrt{2}}{2}=4+3\sqrt{2}\)
\(\Rightarrow T_{min}=4+3\sqrt{2}\) khi \(a=b=\frac{1}{\sqrt{2}}\)