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Áp dụng bất đẳng thức Bunyakovsky, ta được: \(\Sigma_{cyc}\frac{ab}{a^2+bc+ca}=\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Ta có: \(\Sigma_{cyc}\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}=\frac{ab^3+bc^3+ca^3+2.a\sqrt{ab}.c\sqrt{ab}+2.a\sqrt{bc}.b\sqrt{bc}+2.c\sqrt{ca}.b\sqrt{ca}}{\left(ab+bc+ca\right)^2}\le\frac{ab^3+bc^3+ca^3+a^3b+abc^2+a^2bc+b^3c+c^3a+ab^2c}{\left(ab+bc+ca\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}\)
Đẳng thức xảy ra khi a = b = c
Áp dụng BĐT Bunhiacopxki:
\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow\frac{ab}{a^2+bc+ca}\le\frac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)
Tương tự: \(\frac{bc}{b^2+ca+ab}\le\frac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\) ; \(\frac{ac}{c^2+ab+bc}\le\frac{ac\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)
Cộng vế với vế:
\(VT\le\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{ab^3+bc^3+ca^3+2.a\sqrt{ab}.c\sqrt{ab}+2a\sqrt{bc}.b\sqrt{bc}+2c\sqrt{ac}.b\sqrt{ac}}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{ab^3+bc^3+ca^3+a^3b+abc^2+b^3c+a^2bc+ac^3+ab^2c}{\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}\)
\(VT\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta chứng minh bất đẳng thức: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) (a,b,c,x,y,z dương) (Hệ quả của bất đẳng thức Cauchy-Schwarz (Bunyakovsky))
\(\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\sqrt{y}^2+\sqrt{z^2}\right]\ge a^2+b^2+c^2\)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Ta có:
\(A=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(2A=\frac{2bc}{a^2+2bc}+\frac{2ca}{b^2+2ac}+\frac{2ab}{c^2+2ab}\)
\(=\frac{a^2+2bc-a^2}{a^2+2bc}+\frac{b^2+2ca-b^2}{b^2+2ac}+\frac{c^2+2ab-c^2}{c^2+2ab}\)
\(=3-\left(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\right)\)
\(\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=3-1=2\)
=> A<=1
a,b,c dương
Ta viết lại BĐT thành: \(\frac{1}{\frac{a^2}{bc}+2}+\frac{1}{\frac{b^2}{ca}+2}+\frac{1}{\frac{c^2}{ab}+2}\le1\)
Đặt \(\frac{a^2}{bc}=x;\frac{b^2}{ca}=y;\frac{c^2}{ab}=z\Rightarrow\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta cần chứng minh \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\le1\)
Xét biểu thức\(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\) \(\frac{\left(y+2\right)\left(z+2\right)+\left(z+2\right)\left(x+2\right)+\left(x+2\right)\left(y+2\right)}{\left(x+2\right)\left(y+2\right)\left(z+2\right)}\)
\(=\frac{\left(yz+2y+2z+4\right)+\left(zx+2z+2x+4\right)+\left(xy+2x+2y+4\right)}{\left(xy+2x+2y+4\right)\left(z+2\right)}\)
\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+2\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+\left(xy+yz+zx\right)+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(\le\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+3\sqrt{\left(xyz\right)^2}+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}=1\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c
\(\frac{4}{a^2+b^2+c^2}+\frac{2021}{ab+bc+ac}=\frac{4}{a^2+b^2+c^2}+\frac{4}{ab+bc+ac}+\frac{4}{ab+bc+ac}+\frac{2013}{ab+bc+ac}\)
\(=4\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}\right)+\frac{2013}{ab+bc+ac}\)
\(\ge\frac{36}{\left(a+b+c\right)^2}+\frac{2013}{ab+bc+ac}\ge\frac{36}{\left(a+b+c\right)^2}+\frac{2013}{\frac{\left(a+b+c\right)^2}{3}}\ge4+671=675\)
\("="\Leftrightarrow a=b=c=1\)