cho a, b, c >0 thỏa mãn a+b+c=3. CMR \(\dfrac{a}{\sqrt{b+1}}+\dfrac{b}{\sqrt{c+1}}+\dfrac{c}{\sqrt{a+1}}\)>=\(\dfrac{3}{2}\)
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>0 thỏa mãn a+b+c=1. CMR: \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{3}{2}\)
Cho a,b,c>0 thỏa mãn\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\). CMR
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\)
Áp dụng BĐT BSC:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
\(=\dfrac{a+b+c}{2}\)
\(\ge\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
cho a,b dương và c ≠ 0 thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\). CMR: \(\sqrt{a+b}=\sqrt{b+c}+\sqrt{c+a}\)
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b
Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0
Khi đó:
(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)
=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2
=a+b+2c+2|c|=a+b+2c+2|c|
Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c
Do đó:
(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b
⇒√a+c+√b+c=√a+b
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 3 số thực dương a,b.c thỏa mãn abc=1 cmr:\(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Cho 3 số dương a,b,c thỏa mãn
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\)
CMR:\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Đặt vế trái BĐT cần chứng minh là P
Ta có:
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2011}\)
Đồng thời: \(\left\{{}\begin{matrix}y^2+z^2-x^2=2a^2\\z^2+x^2-y^2=2b^2\\x^2+y^2-z^2=2c^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{z^2+x^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)
\(\Rightarrow P\ge\dfrac{1}{2\sqrt{2}}\left(\dfrac{\left(y+z+z+x+x+y\right)^2}{2x+2y+2z}-\left(x+y+z\right)\right)=\dfrac{1}{2\sqrt{2}}\left(x+y+z\right)=\dfrac{1}{2}\sqrt{\dfrac{2011}{2}}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\sqrt{\dfrac{2011}{2}}\)
Cho a,b,c>0 thỏa mãn ab+bc+ac<=1
CMR: \(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(ab+bc+ca\le1\)
\(\Rightarrow\sqrt{a^2+1}\ge\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)
\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{\sqrt{a^2+1}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}+\dfrac{\dfrac{b}{a+b}+\dfrac{b}{b+c}}{2}+\dfrac{\dfrac{c}{b+c}+\dfrac{c}{a+c}}{2}=\dfrac{3}{2}\left(đpcm\right)\)
\(dấu"="\Leftrightarrow a=b=c=\sqrt{\dfrac{1}{3}}\)
1)Cho a;b;c>0 thỏa \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4\)
Chứng minh \(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le1\)
2) Cho a;b;c>0
CMR \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Cho a;b;c>0 thỏa a+b+c=3
CMR \(\dfrac{a+b}{\sqrt{a^2+b^2+6c}}+\dfrac{b+c}{\sqrt{b^2+c^2+6a}}+\dfrac{c+a}{\sqrt{c^2+a^2+6b}}>2\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
cho 3 số thực dương a,b,c thỏa mãn \(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}=2\) .Chứng minh:
\(\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\ge\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\)