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Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
là các số thực không âm, khi đó với 
và 
thì:
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3b\left(b+2\right)}{27b\left(c+2\right)}}=a\)
Tương tự: \(\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c}{3}+\dfrac{a+2}{9}\ge b\)
\(\dfrac{c^3}{a\left(b+2\right)}+\dfrac{a}{3}+\dfrac{b+2}{9}\ge c\)
Cộng vế:
\(VT+\dfrac{4\left(a+b+c\right)}{9}+\dfrac{2}{3}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{5\left(a+b+c\right)}{9}-\dfrac{2}{3}\ge\dfrac{15}{9}-\dfrac{2}{3}=1\)
BT2: Nhân 2 lên, chuyển vế, biến đổi bla..... sẽ ra đpcm
\(\Leftrightarrow\left(1+abc\right)\left(\dfrac{1}{a\left(1+b\right)}+\dfrac{1}{b\left(1+c\right)}+\dfrac{1}{c\left(1+a\right)}\right)\ge3\)
Ta có:
\(\left(1+abc\right).\dfrac{1}{a\left(1+b\right)}=\dfrac{1+abc}{a+ab}=\dfrac{1+a+ab+abc-a-ab}{a+ab}=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}-1\)
\(\Rightarrow VT=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}+\dfrac{1+b}{b\left(1+c\right)}+\dfrac{c\left(1+a\right)}{1+c}+\dfrac{1+c}{c\left(1+a\right)}+\dfrac{a\left(1+b\right)}{1+a}-3\)
\(VT\ge6\sqrt[6]{\dfrac{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}-3=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Giải:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)
\(\Leftrightarrow\) \(\dfrac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)
Do a+1 ; b+1; c+1 >0
\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3
\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0
Áp dụng BĐT Cô-si cho 3 số
Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)
ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3
\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)
\(\Rightarrow\) (*) được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1
Áp dụng BĐT Holder ta có:
\(VT=\left(a+bc\right)\left(\frac{b}{2}+2ac\right)\left(\frac{c}{3}+3ab\right)\)
\(\ge\left(\sqrt[3]{a\cdot\frac{b}{2}\cdot\frac{c}{3}}+\sqrt[3]{bc\cdot2ac\cdot3ab}\right)^3\)
\(=\left(\sqrt[3]{\frac{abc}{6}}+\sqrt[3]{6\left(abc\right)^2}\right)^3\)
\(\ge\left(\sqrt[3]{\frac{6}{6}}+\sqrt[3]{6\cdot6^2}\right)^3=\left(1+6\right)^3=343\)