Cho a,b,c >0 thỏa mãn abc = 1 CMR :
\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\)
Cho các số thực dương a,b,c thỏa mãn abc =1 .CMR
\(\dfrac{3+a}{\left(1+a\right)^2}+\dfrac{3+b}{\left(1+b\right)^2}+\dfrac{3+c}{\left(1+c\right)^2}\ge3\)
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
Cho các số thực dương a,b,c thỏa mãn abc =1 .CMR
\(\dfrac{3+a}{\left(1+a\right)^2}+\dfrac{3+b}{\left(1+b\right)^2}+\dfrac{3+c}{\left(1+c\right)^2}\ge3\)
Cho các số thực dương a,b,c thỏa mãn abc=1 CMR:
\(\dfrac{3+a}{\left(a+1\right)^2}+\dfrac{3+b}{\left(1+b\right)^2}+\dfrac{3+c}{\left(1+c\right)^2}\ge3\)
cho a,b,c thực dương thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le16\left(a+b+c\right)\)
CMR:
\(\dfrac{1}{\left(a+b+2\sqrt{a+c}\right)^3}+\dfrac{1}{\left(b+c+2\sqrt{b+a}\right)^3}+\dfrac{1}{\left(c+a+2\sqrt{c+b}\right)^3}\le\dfrac{8}{9}\)
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
em cũng nghĩ thế mới dùng đc BDT AM-GM 3 số đúng ko thầy :)
Cho abc=1
CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\)
\(VT=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{2}{\left(a+1\right)^2}+\dfrac{2}{\left(b+1\right)^2}+\dfrac{2}{\left(c+1\right)^2}\)
Mặt khác:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1.1\right)^2}\ge\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{a}{b}\right)}+\dfrac{1}{\left(1+ab\right)\left(1+\dfrac{b}{a}\right)}=\dfrac{1}{1+ab}\)
Do đó:
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\)
\(VT\ge\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}+\dfrac{1}{1+\dfrac{1}{c}}+\dfrac{1}{1+\dfrac{1}{a}}+\dfrac{1}{1+\dfrac{1}{b}}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 1: Cho a, b, c > 1. CMR: \(a^{\log_bc}+b^{\log_ca}+c^{\log_ab}\ge3\sqrt[3]{abc}\)
Bài 2: Cho các số x, y, z > 0 thoả mãn: \(\dfrac{x\left(y+z-x\right)}{logx}=\dfrac{y\left(z+x-y\right)}{logy}=\dfrac{z\left(x+y-z\right)}{logz}\). CMR: xy.yx = yz.zy = xz.zx
Cho a,b,c>0 thỏa mãn ab+bc+ca=1. CMR:
\(\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^3\le\dfrac{3}{2}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
Đẳng thức quen thuộc: \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\) và tương tự cho các mẫu số còn lại
Ta có:
\(\sum\dfrac{1}{a^2+1}=\sum\dfrac{1}{\left(a+b\right)\left(a+c\right)}=\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Mặt khác:
\(2\left(ab+bc+ca\right)\left(a+b+c\right)=\left[a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right]\left(a+b+c\right)\)
\(\ge\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\) (Bunhiacopxki)
\(\Rightarrow\sum\dfrac{1}{a^2+1}\ge\dfrac{\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\left(\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)^2\)
\(=\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^2\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{3}{2}\)
Đúng theo AM-GM:
\(\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Cho a , b , c dương
CMR \(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\right]^4}\)
\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\\\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}\end{matrix}\right.\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge1+3\sqrt[3]{\dfrac{1}{abc}}+3\sqrt[3]{\dfrac{1}{a^2b^2c^2}}+\dfrac{1}{abc}\)
\(\Rightarrow1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(\Rightarrow3\left(\sqrt[3]{1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{abc}}\right)^4\ge3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\) (2)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt[3]{abc}\le\dfrac{abc+1+1}{3}=\dfrac{abc+2}{3}\)
\(\Rightarrow1+\dfrac{1}{\sqrt[3]{abc}}\ge1+\dfrac{3}{abc+2}\)
\(\Rightarrow3\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) (3)
Từ (1) và (2) và (3)
\(\Rightarrow VT\ge3\left(1+\dfrac{3}{abc+2}\right)^4\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{abc+2}\right)^4\) ( đpcm )
cho các số dương a,b,c sao cho abc=1 . chứng minh :
\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\)
Tham khảo:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24