Question

[❥一Ǥoɗɗεşş øf iηէεℓℓεcէ︵♛╾━╤デ╦︻]

Cho a, b, c là các số thực dương CMR:\(a+b+c\ge\frac{4ab}{4ab+1}+\frac{4bc}{4bc+1}+\frac{4ca}{4ca+1}\)

Answer 1

[☆MĭηɦღAηɦ❄]

Ta có : 

\(\frac{4ab+1}{4ab}=1+\frac{1}{4ab}\ge1+\frac{1}{\left(a+b\right)^2}\)

\(\Rightarrow\frac{4ab}{4ab+1}\le\frac{1}{1+\frac{1}{\left(a+b\right)^2}}\)

Tương tự ta được : 

\(\frac{4bc}{4bc+1}\le\frac{1}{1+\frac{1}{\left(b+c\right)^2}};\frac{4ca}{4ca+1}\le\frac{1}{1+\frac{1}{\left(c+a\right)^2}}\)

\(\Rightarrow VP\le\frac{1}{1+\frac{1}{\left(a+b\right)^2}}+\frac{1}{1+\frac{1}{\left(b+c\right)^2}}+\frac{1}{1+\frac{1}{\left(c+a\right)^2}}\)

BĐT cần chứng minh tương đương với 

\(a+b+c\ge\frac{1}{1+\frac{1}{\left(a+b\right)^2}}+\frac{1}{1+\frac{1}{\left(b+c\right)^2}}+\frac{1}{1+\frac{1}{\left(c+a\right)^2}}\) (1)

Đặt \(a+b=x;b+c=y;c+a=z\)

\(x,y,z>0;x+y+z=2\left(a+b+c\right)\)

\(\Rightarrow\left(1\right)\Leftrightarrow x+y+z\ge2\left(\frac{1}{1+\frac{1}{x^2}}+\frac{1}{1+\frac{1}{y^2}}+\frac{1}{1+\frac{1}{z^2}}\right)\)

\(VP=\frac{2x^2}{x^2+1}+\frac{2y^2}{y^2+1}+\frac{2z^2}{z^2+1}\le\frac{2x^2}{2x}+\frac{2y^2}{2y}+\frac{2z^2}{2z}=x+y+z=VT\)

Vậy BĐT được chứng minh

Dấu "=" xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c=\frac{1}{2}\)

Answer 2

[yen dang]

\(\frac{4ab}{4ab+1}< =\frac{4ab}{2\sqrt{4ab}}=\sqrt{ab}\)

CMTT =>\(\hept{\begin{cases}\frac{4bc}{4bc+1}< =\sqrt{bc}\\\frac{4ac}{4ac+1}< =\sqrt{ac}\end{cases}}\)

Ta có \(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ac}\)

=\(\frac{1}{2}\left(\left(a+2\sqrt{ab}+b\right)+\left(b+2\sqrt{bc}+c\right)+\left(c+2\sqrt{ac}+a\right)\right)\)

=\(\frac{1}{2}\left(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\right)>=0\)

dấu = xảy ra khi a=b=c.

\(=>a+b+c>=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)\(>=\frac{4ab}{4ab+1}+\frac{4bc}{4bc+1}+\frac{4ac}{4ac+1}\)