Bài 1: Cho \(a,b,c>0;a+b+c=9\). CMR:
\(\frac{a^3+b^3}{ab+9}+\frac{b^3+c^3}{bc+9}+\frac{c^3+a^3}{ca+9}\ge9\)
Bài 2: Cho \(a,b,c>0;a^2+b^2+c^2=3\). CMR:
\(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\ge\frac{a+b+c}{2}\)
Akai Haruma đây nha chị :D
Bài 1:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^4}{a^2b+9a}+\frac{b^4}{ab^2+9b}+\frac{b^4}{b^2c+9b}+\frac{c^4}{bc^2+9c}+\frac{c^4}{c^2a+9c}+\frac{a^4}{ca^2+9a}\)
\(\ge \frac{(a^2+b^2+b^2+c^2+c^2+a^2)^2}{ab(a+b)+bc(b+c)+ca(c+a)+18(a+b+c)}=\frac{4(a^2+b^2+c^2)^2}{ab(a+b)+bc(b+c)+ca(c+a)+162}\)
Áp dụng BĐT AM-GM:
\(a^3+b^3+c^3=\frac{a^3+b^3+b^3}{3}+\frac{b^3+c^3+c^3}{3}+\frac{c^3+a^3+a^3}{3}\geq ab^2+bc^2+ca^2\)
Tương tự: \(a^3+b^3+c^3\geq a^2b+b^2c+c^2a\)
\(\Rightarrow a^3+b^3+c^3\geq \frac{ab(a+b)+bc(b+c)+ca(c+a)}{2}\)
\(\Rightarrow a^3+b^3+c^3+ab(a+b)+bc(c+a)+ca(c+a)\geq \frac{3}{2}[ab(a+b)+bc(b+c)+ca(c+a)]\)
\(\Leftrightarrow (a^2+b^2+c^2)(a+b+c)\geq \frac{3}{2}[ab(a+b)+bc(b+c)+ca(c+a)]\)
\(\Leftrightarrow ab(a+b)+bc(b+c)+ca(c+a)\leq 6(a^2+b^2+c^2)\)
Do đó: \(\text{VT}\geq \frac{4(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)+162}\)
Đặt \(a^2+b^2+c^2=t\). Dễ thấy \(t\geq \frac{(a+b+c)^2}{3}=27\). Khi đó:
\(\frac{4(a^2+b^2+c^2)^2}{6(a^2+b^2+c^2)+162}-9=\frac{4t^2}{6t+162}-9=\frac{2(t-27)(2t+27)}{6t+162}\geq 0, \forall t\geq 27\)
\(\Rightarrow \text{VT}\geq \frac{4t^2}{6t+162}\geq 9\) (đpcm). Dấu "=" xảy ra khi $a=b=c=3$
Bài 2:
Áp dụng BĐT AM-GM:
\(\text{VT}=a-\frac{ab^2}{a+b^2}+b-\frac{bc^2}{b+c^2}+c-\frac{ca^2}{c+a^2}=(a+b+c)-\left(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\right)\)
\(\geq (a+b+c)-\left(\frac{ab^2}{2\sqrt{ab^2}}+\frac{bc^2}{2\sqrt{bc^2}}+\frac{ca^2}{\sqrt{ca^2}}\right)=(a+b+c)-\frac{1}{2}(\sqrt{ab^2}+\sqrt{bc^2}+\sqrt{ca^2})\)
\(\geq (a+b+c)-\frac{1}{2}\left(\frac{ab+b}{2}+\frac{bc+c}{2}+\frac{ca+a}{2}\right)=\frac{3(a+b+c)-(ab+bc+ac)}{2}\)
Tiếp tục áp dụng BĐT AM-GM:
\((a+b+c)^2\geq 3(ab+bc+ac)=(a^2+b^2+c^2)(ab+bc+ac)\geq (ab+bc+ac)^2\)
\(\Rightarrow a+b+c\geq ab+bc+ac\)
Do đó: \(\text{VT}\geq \frac{3(a+b+c)-(a+b+c)}{2}=\frac{a+b+c}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$