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Ta có:
\(\left(a^2+1\right)+\left(b^2+1\right)+\left(c^2+1\right)+\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\)
\(\ge2a+2b+2c+2ab+2bc+2ca=12\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)
\(\Rightarrow a^2+b^2+c^2\ge3\)
\(P=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}\)
\(P\ge a^2+b^2+c^2\ge3\)
\(P_{min}=3\) khi \(a=b=c=1\)
Đề đúng: Cho a,b,c thỏa mãn a+b+c>0; ab+bc+ac>0; abc>0. Chứng minh a,b,c>0
Vì abc>0 nên có ít nhất 1 số lớn hơn 0
Vai trò của a, b, c như nhau nên chọn a>0
TH1: b<0;c<0
\(\Rightarrow b+c>-a\Rightarrow\left(b+c\right)^2< -a\left(b+c\right)\)
\(\Rightarrow b^2+2bc+c^2< -ab-ac\)
\(\Rightarrow b^2+bc+c^2< -\left(ab+bc+ca\right)\)(vô lí)
TH2: b>0, c>0 thì a>0( luôn đúng)
Vậy a, b, c >0
a: \(\Leftrightarrow\left(a+1\right)^2-4a\ge0\)
hay \(\left(a-1\right)^2>=0\)(luôn đúng)
b: \(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(c^2+d^2\right)\left(a^2+b^2\right)=VP\)
Đặt vế trái là P
\(P=\dfrac{1.c+ab}{a+b}+\dfrac{1.a+bc}{b+c}+\dfrac{1.b+ac}{a+c}=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)
\(P=\dfrac{ac+c^2+bc+ab}{a+b}+\dfrac{a^2+ac+ab+bc}{b+c}+\dfrac{ab+ac+b^2+bc}{a+c}\)
\(P=\dfrac{c\left(a+c\right)+b\left(a+c\right)}{a+b}+\dfrac{a\left(a+c\right)+b\left(a+c\right)}{b+c}+\dfrac{a\left(b+c\right)+b\left(b+c\right)}{a+c}\)
\(P=\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)
Áp dụng BĐT Cô-si:
\(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)}}=2\left(a+c\right)\) (1)
Tương tự: \(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(b+c\right)\) (2)
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\) (3)
Cộng vế với vế (1);(2);(3):
\(2.\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+2.\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+2.\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+b\right)+2\left(b+c\right)+2\left(c+a\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+c}\ge2\left(a+b+c\right)=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)