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\(VT=\frac{ab}{ab+c}+\frac{ac}{ac+b}+\frac{bc}{bc+a}\)
\(=\frac{ab}{ab+\left(a+b+c\right)c}+\frac{ac}{ac+\left(a+b+c\right)b}+\frac{bc}{bc+\left(a+b+c\right)a}\)
\(=\frac{ab}{\left(b+c\right)\left(c+a\right)}+\frac{ac}{\left(a+b\right)\left(b+c\right)}+\frac{bc}{\left(a+b\right)\left(c+a\right)}\)
\(=\frac{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Cần chứng minh \(\frac{ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{3}{4}\)
\(\Leftrightarrow a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\ge6abc\)
BĐT cuối luôn đúng theo AM-GM
Áp dụng bđt AM-GM :
\(\frac{1}{a^2+1}+\frac{a^2+1}{4}\ge2\sqrt{\frac{a^2+1}{\left(a^2+1\right)\cdot4}}=1\)
Tương tự ta có :
\(\frac{1}{b^2+1}+\frac{b^2+1}{4}\ge1\)
\(\frac{1}{c^2+1}+\frac{c^2+1}{4}\ge1\)
Cộng từng vế ta có :
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}+\frac{a^2+b^2+c^2+3}{4}\ge3\)
Áp dụng bđt quen thuộc : \(a^2+b^2+c^2\ge ab+bc+ac=3\)
Khi đó : \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge3-\frac{3+3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
bạn làm sai rồi . Khi \(a^2+b^2+c^2\ge3\) bạn chuyển vế thì nó không cùng dấu với bất đẳng thức
Bài làm :
\(VT=\frac{a}{b\left(b^2+a\right)}+\frac{b}{c\left(c^2+b\right)}+\frac{c}{a\left(a^2+c\right)}\)
\(=\frac{1}{b}\cdot\frac{a}{b^2+a}+\frac{1}{c}\cdot\frac{b}{c^2+b}+\frac{1}{a}\cdot\frac{c}{a^2+c}\)
\(=\frac{1}{b}\cdot\left(1-\frac{b^2}{b^2+a}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{c^2+b}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{a^2+c}\right)\)
Áp dụng BĐT Cô-si :
\(VT\ge\frac{1}{b}\cdot\left(1-\frac{b^2}{2b\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c^2}{2c\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a^2}{2a\sqrt{c}}\right)\)
\(=\frac{1}{b}\cdot\left(1-\frac{b}{2\sqrt{a}}\right)+\frac{1}{c}\cdot\left(1-\frac{c}{2\sqrt{b}}\right)+\frac{1}{a}\cdot\left(1-\frac{a}{2\sqrt{c}}\right)\)
\(=\frac{1}{b}-\frac{1}{2\sqrt{a}}+\frac{1}{c}-\frac{1}{2\sqrt{b}}+\frac{1}{a}-\frac{1}{2\sqrt{c}}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)
Lại áp dụng BĐT Cô-si :
\(\frac{1}{\sqrt{a}}\le\frac{\frac{1}{a}+1}{2};\frac{1}{\sqrt{b}}\le\frac{\frac{1}{b}+1}{2};\frac{1}{\sqrt{c}}\le\frac{\frac{1}{c}+1}{2}\)
Do đó :
\(VT\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{2}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}{2}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)
\(=\frac{3}{4}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\ge\frac{3}{4}\cdot\frac{9}{a+b+c}-\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
giúp vs
Lê Thị Thục HiềnTrần Thanh PhươngVũ Minh Tuấn
\(\frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{bc^2+abc+ba^2}+\frac{c^2}{ca^2+abc+cb^2}\) (1)
Áp dụng BDT Cauchy-Schwarz: \(\left(1\right)\ge\frac{\left(a+b+c\right)^2}{ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc}\)
Lại có: \(ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+3abc=\left(ab+bc+ac\right)\left(a+b+c\right)\)
Thay vào -> dpcm
\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)
Áp dụng BĐT Cauchy dạng phân thức
\(\Rightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}\)
\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu "=" xảy ra khi a=b=c
Chúc bạn học tốt !!!
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