4.
Ta có:
\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+ab+b^2}}=\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+ab+b^2\right)}}\)
\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+2ab+b^2+2c^2}\ge\dfrac{2\left(ab+2c^2\right)}{2a^2+2b^2+2c^2}=\dfrac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự và cộng lại:
\(P\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
5.
Ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{a^2+b^2+c\left(a+b\right)}{\left(b+c\right)\left(c+a\right)}\ge\dfrac{\left(a+b\right)^2+2c\left(a+b\right)}{2\left(b+c\right)\left(c+a\right)}=\dfrac{\left(a+c+b+c\right)\left(a+b\right)}{2\left(b+c\right)\left(c+a\right)}\)
Tương tự...
Đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\)
\(\Rightarrow P\ge\dfrac{x\left(y+z\right)}{2yz}.\dfrac{y\left(z+x\right)}{2zx}.\dfrac{z\left(x+y\right)}{2xy}=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8xyz}\)
\(\Rightarrow P\ge\dfrac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{8xyz}=1\)
3.
Ta có:
\(\dfrac{\left|a-b\right|}{\sqrt{c^2+2ab}}=\dfrac{\left(a-b\right)^2}{\sqrt{\left(a-b\right)^2\left(c^2+2ab\right)}}\ge\dfrac{2\left(a-b\right)^2}{\left(a-b\right)^2+c^2+2ab}=\dfrac{2\left(a-b\right)^2}{a^2+b^2+c^2}\)
Tương tự và cộng lại:
\(P\ge\dfrac{2\left(a-b\right)^2+2\left(b-c\right)^2+2\left(c-a\right)^2}{a^2+b^2+c^2}=\dfrac{4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}{a^2+b^2+c^2}\)
\(=\dfrac{4\left(a^2+b^2+c^2\right)-2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\)
2.
\(P=\sqrt{\dfrac{1}{1+8\left(\dfrac{y}{x}\right)^3}}+\sqrt{\dfrac{4\left(\dfrac{y}{x}\right)^3}{\left(\dfrac{y}{x}\right)^3+\left(\dfrac{y}{x}+1\right)^3}}\)
Đặt \(\dfrac{y}{x}=a\) \(\Rightarrow P=\sqrt{\dfrac{1}{8a^3+1}}+\sqrt{\dfrac{4a^3}{a^3+\left(a+1\right)^3}}\)
Ta có:
\(\dfrac{1}{\sqrt{8a^3+1}}=\dfrac{1}{\sqrt{\left(2a+1\right)\left(4a^2-2a+1\right)}}\ge\dfrac{2}{4a^2+2}=\dfrac{1}{2a^2+1}\)
Ta sẽ chứng minh:
\(\sqrt{\dfrac{4a^3}{a^3+\left(a+1\right)^3}}\ge\dfrac{2a^2}{2a^2+1}\)
Thật vậy, BĐT tương đương:
\(\left(2a^2+1\right)^2\ge a\left(2a^3+3a^2+3a+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2\left(2a^2+a+1\right)\ge0\) (luôn đúng)
Do đó:
\(P\ge\dfrac{1}{2a^2+1}+\dfrac{2a^2}{2a^2+1}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=1\) hay \(x=y\)
1.
Đặt \(\left(a;b;c\right)=\left(\dfrac{x^3}{8};\dfrac{y^3}{8};\dfrac{z^3}{8}\right)\Rightarrow xyz=8\)
BĐT trở thành:
\(\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}\ge1\)
Ta có:
\(\dfrac{1}{\sqrt{x^3+1}}=\dfrac{1}{\sqrt{\left(x+1\right)\left(x^2-x+1\right)}}\ge\dfrac{2}{x^2+2}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{2}{x^2+2}+\dfrac{2}{y^2+2}+\dfrac{2}{z^2+2}\ge1\)
\(\Leftrightarrow x^2+y^2+z^2\ge12\) (quy đồng rút gọn)
BĐT trên đúng do \(x^2+y^2+z^2\ge3\sqrt[3]{\left(xyz\right)^2}=3\sqrt[3]{8^2}=12\)
Dấu "=" xảy ra khi \(a=b=c=1\)
