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\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ac}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Áp dụng BĐT cosi
\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3}{4}a\)
Tương tự
=> \(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{1}{4}\left(a+b+c\right)\)
Lại có \(\left(a+b+c\right)\ge\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{9}{1}=9\)
=> \(A\ge\frac{9}{4}\)
MinA=9/4 khi a=b=c=3
Ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.1=3\) \(\Rightarrow a+b+c\ge\sqrt{3}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel
\(B=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{3}}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\\ab+bc+ca=1\end{cases}}\) \(\Leftrightarrow\) \(a=b=c=\frac{\sqrt{3}}{3}\)
\(P=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{2016}{ab+bc+ca}\)
\(P\ge\frac{9}{a^2+b^2+c^2+ab+bc+ca+ab+bc+ca}+\frac{2016}{\frac{1}{3}\left(a+b+c\right)^2}\)
\(P\ge\frac{6057}{\left(a+b+c\right)^2}\ge\frac{6057}{3^2}=673\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\) (áp dụng Bất Đẳng Thức Cosi)
\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)
\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)
Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)
Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)
Dấu "=" xảy ra khi a=b=c
\(P=\frac{2018}{a^2+b^2+c^2}+\frac{2018}{ab+bc+ac}-\frac{2017}{a^2+b^2+c^2}\)
\(P\ge2018\left(\frac{4}{a^2+b^2+c^2+ab+bc+ac}\right)-\frac{2017}{a^2+b^2+c^2}\)
\(P\ge\frac{2018.8}{\left(a+b+c\right)^2}-\frac{2017}{a^2+b^2+c^2}=\frac{2018.8}{9}-\frac{2017}{a^2+b^2+c^2}\)
Vì \(9=\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2\ge3\)
\(P\ge\frac{2018.8}{9}-\frac{2017}{3}=...\)
P min = ... khi a=b=c = 1