\(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(1+a^2\right)\left(1+b^2\right)}\le\frac{1}{2}\)
Cho a,b là số thực
CMR: \(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\\ \)
Ta có: \(\left(\left|x\right|-\left|y\right|\right)^2\ge0\)
\(\Rightarrow x^2+y^2\ge2\left|xy\right|\)
\(\Rightarrow\left|\frac{2xy}{x^2+y^2}\right|\le1\)(*)
Lại có: \(\left(a+b\right)^2+\left(1-ab\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
Nên: \(\left|\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\right|=\left|\frac{\left(a+b\right)\left(1-ab\right)}{\left(a+b\right)^2+\left(1-ab\right)^2}\right|\)
Áp dụng (*), ta có: \(\left|\frac{\left(a+b\right)\left(1-ab\right)}{\left(a+b\right)^2+\left(1-ab\right)^2}\right|\le\frac{1}{2}\)
\(\Rightarrow\left|\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\right|\le\frac{1}{2}\)
\(\Rightarrow\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\) \(\left(đpcm\right)\)
CMR: \(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\)
cmr
\(-\frac{1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(1+a^2\right)\left(1+b^2\right)}\le\frac{1}{2}\)
Sử dụng bất đẳng thức quen thuộc: \(4ab\le\left(a+b\right)^2\)
Ta có:
\(\Rightarrow\left[\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\right]^2\le\frac{1}{4}\)
\(\Rightarrow\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\)
Vậy \(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\le\frac{1}{2}\left(đpcm\right)\)
Chứng minh: \(\frac{-1}{2}\le\frac{\left(a+b\right) \left(1-ab\right)}{\left(a^2 +1\right)\left(b^2+1\right)}\le\frac{1}{2}\)
Ta chứng minh
\(\frac{-1}{2}\le\frac{\left(a+b\right)\left(1-ab\right)}{\left(a^2+1\right)\left(b^2+1\right)}\)
\(\Leftrightarrow2\left(a+b\right)\left(1-ab\right)+\left(a^2+1\right)\left(b^2+1\right)\ge0\)
\(\Leftrightarrow\left(ab-a-b-1\right)^2\ge0\)(đúng)
Tương tự cho trường hợp còn lại ta có ĐPCM
a, Cho a,b là các số thực dương và ab<1. Chứng minh \(\frac{1}{1+a}+\frac{1}{1+b}\le\frac{2}{1+\sqrt{ab}}\)
b, Cho a,b,c là các số thực dương thỏa mãm abc=1. Chứng minh:
\(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)
Cho a; b; c là các số thực dương thỏa mãn ab + bc + ca = 3.
CMR: \(\frac{1}{1+a^2\left(b+c\right)}+\frac{1}{1+b^2\left(c+a\right)}+\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{abc}\)
Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
đặt \(P=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\)
\(\Rightarrow P-3=\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\le\frac{ab}{1-\frac{a^2+b^2}{2}}+\frac{bc}{1-\frac{b^2+c^2}{2}}+\frac{ca}{1-\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{2}.\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{1}{2}.\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(c^2+a^2\right)}+\frac{1}{2}.\frac{\left(c+a\right)^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\)
\(\le\frac{1}{2}.\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+a^2}+\frac{c^2}{b^2+c^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{2}\)
\(\Rightarrow P-3\le\frac{3}{2}\Rightarrow P\le\frac{9}{2}\)
cho đề này:
cho a;b;c là các số thực dương thỏa mãn a2+b2+c2=1.CMR:\(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\le\frac{9}{2}\)
cho \(0< a\le\frac{1}{2},0< b\le\frac{1}{2}.CM:\left(\frac{a+b}{2-a-b}\right)^2\ge\frac{ab}{\left(1-a\right)\left(1-b\right)}\)
\(\left(\frac{a+b}{2-a-b}\right)^2\ge\frac{ab}{\left(1-a\right)\left(1-b\right)}\)
\(\Leftrightarrow\left(\frac{a+b}{2-a-b}\right)^2-\frac{ab}{\left(1-a\right)\left(1-b\right)}\ge0\)
\(\Leftrightarrow\frac{\left(a^2+2ab+b^2\right)\left(a-1\right)\left(b-1\right)-ab\left(a+b-2\right)^2}{\left(a+b-2\right)^2\left(a-1\right)\left(b-1\right)}\ge0\)
\(\Leftrightarrow\frac{-a^3-b^3+a^2+b^2+a^2b+ab^2-2ab}{\left(a+b-2\right)^2\left(a-1\right)\left(b-1\right)}\ge0\)
\(\Leftrightarrow\frac{-\left(a-b\right)^2\left(a+b-1\right)}{\left(a+b-2\right)^2\left(a-1\right)\left(b-1\right)}\ge0\)
BĐT cuối luôn đúng vì \(a;b\in\)\((0;\frac{1}{2}]\)
cho a, b, c là ba số thực dương.CM : \(\frac{1}{a\left(a^2+8bc\right)}+\frac{1}{b\left(b^2+8ca\right)}+\frac{1}{c\left(c^2+8ab\right)}\le\frac{1}{3abc}\)
đề bài \(\Leftrightarrow\frac{bc}{a^2+8bc}+\frac{ca}{b^2+8ca}+\frac{ab}{c^2+8ab}\le\frac{1}{3}\)
\(\Leftrightarrow\left(\frac{1}{8}-\frac{bc}{a^2+8bc}\right)+\left(\frac{1}{8}+\frac{ca}{b^2+8ca}\right)+\left(\frac{1}{8}-\frac{ab}{c^2+8ab}\right)\ge\frac{1}{24}\)
\(\Leftrightarrow\frac{a^2}{a^2+8bc}+\frac{b^2}{b^2+8ca}+\frac{c^2}{c^2+8ab}\ge\frac{1}{3}\)
Mặt khác: vế trái \(\frac{a^2}{a^2+8bc}+\frac{b^2}{b^2+8ca}+\frac{c^2}{c^2+8ab}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+8\left(ab+bc+ca\right)}\)\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+6\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+2\left(a+b+c\right)^2}=\frac{1}{3}\)
=> đpcm