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\(P=\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{\left(a+b+c\right)^3}{abc}\)
\(\ge\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{9\left(a+b+c\right)^2}{ab+bc+ca}\)
\(=\left[\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\right]+\frac{8\left(a^2+b^2+c^2\right)}{ab+bc+ca}+18\)
\(\ge2+8+18=28\)
Đẳng thức xảy ra khi \(a=b=c\)
\(2P=\frac{2ab+2bc+2ca}{a^2+b^2+c^2}+\frac{2\left(a+b+c\right)^2}{abc}=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\frac{2\left(a+b+c\right)^3}{abc}\)
\(\Rightarrow2P+1=\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{2\left(a+b+c\right)}{abc}\right)=\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\right)\)
\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{18}{ab+bc+ca}\right)\)
\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{16}{ab+bc+ca}\right)\)
\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{16}{ab+bc+ca}\right)\)
\(\Rightarrow2P+1\ge\left(a+b+c\right)^2\left(\frac{9}{\left(a+b+c\right)^2}+\frac{48}{\left(a+b+c\right)^2}\right)=57\)
\(\Rightarrow P\ge28\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{4abc}\)
\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\ge\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\ge\frac{9}{ab+ac+bc}\right)\)
\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}\ge\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}\right)-\frac{3}{2}\left(1\right)\)
Lại có:\(\frac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2+2\left(ab+bc+ac\right)}{30\left(a^2+b^2+c^2\right)}\)
\(=\frac{1}{30}+\frac{1}{15}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\left(2\right)\).Từ (1);(2) có:
\(P=\frac{1}{30}-\frac{3}{2}+\frac{1}{5}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)+\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)
\(=\frac{1}{15}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}+\frac{ab+bc+ca}{a^2+b^2+c^2}-22\right)\ge-\frac{4}{3}\)
đề thi hsg toán lớp 9 tỉnh thanh hóa năm 2016-2017 mà
Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{4abc}\)
\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\ge\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\ge\frac{9}{ab+ac+bc}\right)\)
\(\Rightarrow\frac{a^3+b^3+c^3}{4abc}\ge\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}\right)-\frac{3}{2}\left(1\right)\)
Lại có: \(\frac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2+2\left(ab+bc+ac\right)}{30\left(a^2+b^2+c^2\right)}\)
\(=\frac{1}{30}+\frac{1}{15}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow P=\frac{1}{30}-\frac{3}{2}+\frac{1}{5}\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)+\frac{9}{4}\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)
\(=\frac{1}{15}\left(\frac{a^2+b^2+c^2}{ab+bc+ac}+\frac{ab+bc+ca}{a^2+b^2+c^2}-22\right)\ge-\frac{4}{3}\)
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