- Áp dụng BĐT Cô-si, ta có:
\(\dfrac{a^2}{1+b}+\dfrac{1+b}{4}\ge2\sqrt{\dfrac{a^2}{1+b}.\dfrac{1+b}{4}}=a\)
- CMTT: \(\dfrac{b^2}{1+c}+\dfrac{1+c}{4}\ge b;\dfrac{c^2}{1+a}+\dfrac{1+a}{4}\ge c\)
- Cộng từng vế của các BĐT trên ta có:
\(\dfrac{a^2}{1+b}+\dfrac{b^2}{1+c}+\dfrac{c^2}{1+a}+\dfrac{a+b+c+3}{4}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{1+b}+\dfrac{b^2}{1+c}+\dfrac{c^2}{1+a}\ge\dfrac{3\left(a+b+c\right)-3}{4}\)
Mà ta có: \(a+b+c\ge3\sqrt[3]{abc}=3\sqrt[3]{1}=3\)
\(\Rightarrow\dfrac{a^2}{1+b}+\dfrac{b^2}{1+c}+\dfrac{c^2}{1+a}\ge\dfrac{3.3-3}{4}=\dfrac{3}{2}\left(đpcm\right)\)
- Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Đặt vế trái là P
\(a+b+c\ge3\sqrt[3]{abc}=3\Rightarrow3+a+b+c\le2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^2}{1+b}+\dfrac{b^2}{1+c}+\dfrac{c^2}{1+a}\ge\dfrac{\left(a+b+c\right)^2}{3+a+b+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: `a^3 + b^3 + c^3 >= 3abc`.
`-> a + b + c >= 3` \(\sqrt[3]{abc}\) `= 3`.
Áp dụng bất đẳng thức Svac - xơ:
`a^2/(1+b) + b^2/(1+c) +c^2/(1+a) >= (a+b+c)^2/(1 + b + 1+c+1+a) = (a+b+c)^2/(3+a+b+c) >= 3^2/(3+3) = 9/6 = 3/2`.
Dấu bằng xảy ra `<=> a = b = c`.
