cho \(a+b+c=1\) và \(a,b,c\ge-\frac{1}{4}\) chứng minh rằng :
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
Cho a. b. c không âm và có tổng bằng 1. Chứng minh rằng:
\(\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
đặt \(S=\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\)
\(=\frac{a^3}{4a^2b^2+a^2}+\frac{b^3}{4b^2c^2+b^2}+\frac{c^3}{4a^2c^2+c^2}\ge\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4a^2b^2+4b^2c^2+4c^2a^2+a^2+b^2+c^2}\)
xét hiệu:
1-4(a2b2+b2c2+c2a2)-a2-b2-c2
=2ab+2bc+2ca-4(a2b2+b2c2+c2a2)
=2ab(1-2ab)+2bc(1-2bc)+2ca(1-2ca)
ta có:
\(2ab\le\frac{\left(a+b\right)^2}{2}\le\frac{1}{2};2bc\le\frac{\left(b+c\right)^2}{2}\le\frac{1}{2};2ca\le\frac{\left(c+a\right)^2}{2}\le\frac{1}{2}\)
\(\Rightarrow2ab\left(1-2ab\right);2bc\left(1-2bc\right);2ca\left(1-2ca\right)\ge0\)
\(\Rightarrow1\ge4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2\)
\(\Rightarrow\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
\(\Rightarrow\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
=>đpcm
dấu"=" xảy ra khi 1 số=1;2 số còn lại =0
Cho a,b,c > \(\dfrac{-1}{4}\). Chứng minh rằng
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}\)
Đề bài thiếu, chắc chắn phải có thêm 1 dữ kiện khác
Ví dụ, bạn cho \(a=b=c=1000\) sẽ thấy BĐT sai
cho a+b+c=1 và \(a,b,c\ge-\frac{1}{4}\) CMR \(\sqrt{4a+1}+\sqrt{4b+1}\sqrt{4c+1}< 5\)
Áp dụng Cauchy-Schwarz:
\(VT^2\le\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)\)
\(=3\left(4\left(a+b+c\right)+3\right)\)
\(=3\left(4+3\right)=21< 25=VP^2\)
Suy ra VT<VP---> đúng
Cho 1< a, b, c <2. Chứng minh rằng
\(\frac{b\sqrt{a}}{4b\sqrt{c}-c\sqrt{a}}+\frac{c\sqrt{b}}{4c\sqrt{a}-a\sqrt{b}}+\frac{a\sqrt{c}}{4a\sqrt{b}-b\sqrt{c}}\ge1\)
Ta có: \(4b\sqrt{c}-c\sqrt{a}=\sqrt{c}\left(4b-\sqrt{ac}\right)>0\)( do \(1< a,b,c< 2\))
Tương tự => Các MS dương
\(VT=\frac{ba}{4b\sqrt{ac}-ca}+\frac{cb}{4c\sqrt{ba}-ab}+\frac{ac}{4a\sqrt{bc}-bc}\)
Áp dụng BĐT cosi schawr ta có
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{4b\sqrt{ac}+4c\sqrt{ab}+4a\sqrt{bc}-ab-bc-ac}\)
Áp dụng cosi ta có \(2b\sqrt{ac}=2\sqrt{ab}.\sqrt{ac}\le ab+ac\);\(2c\sqrt{ab}\le ac+bc\);\(2a\sqrt{bc}\le ab+ac\)
=> \(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{ab+bc+ac+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}}=\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}=1\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
1) Cho a,b,c>0 và a+b+c=3
Chứng minh rằng \(\frac{1}{4a^2+b^2+c^2}+\frac{1}{a^2+4b^2+c^2}+\frac{1}{a^2+b^2+4c^2}\le\frac{1}{2}\)
2) Giaỉ phương trình
\(\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}=16-\sqrt{x-2}-\sqrt{y-1}-\sqrt{z-5}\)
Thôi giải lại câu 1:v (ý tưởng dồn biến là quá trâu bò! Bên AoPS em mới phát hiện ra có một cách bằng Cauchy-Schwarz quá hay!)
\(BĐT\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{9}{2}\)(*)
BĐT này đúng theo Cauchy-Schwarz: \(VT_{\text{(*)}}\le\Sigma_{cyc}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{9}{2}\)
Ta có đpcm.
Equality holds when a = b = c = 1 (Đẳng thức xảy ra khi a = b =c = 1)
1/Đặt \(VT=f\left(a;b;c\right)\) và \(0< t=\frac{a+b}{2}\)
Ta có: \(f\left(a;b;c\right)-f\left(t;t;c\right)=\frac{1}{4a^2+b^2+c^2}+\frac{1}{4b^2+a^2+c^2}-\frac{2}{5t^2+c^2}+\frac{1}{a^2+b^2+4c^2}-\frac{1}{2t^2+4c^2}\)
\(=\frac{5t^2-4a^2-b^2}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}+\frac{5t^2-4b^2-a^2}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)
\(=-\frac{1}{4}\left(a-b\right)\left[\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\right]+\frac{2t^2-a^2-b^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)
Xét cái ngoặc to: \(\frac{\left(11a+b\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{\left(a+11b\right)}{\left(5t^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)
\(=\frac{\left(11a+b\right)\left(4b^2+a^2+c^2\right)-\left(a+11b\right)\left(4a^2+b^2+c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)
\(=\frac{\left(a-b\right)\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}\)
Từ đó: f(a;b;c) -f(t;t;c)
\(=-\frac{\frac{1}{4}\left(a-b\right)^2\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{-\frac{1}{2}\left(a-b\right)^2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\)
\(=-\frac{1}{4}\left(a-b\right)^2\left[\frac{\left(7a^2-36ab+7b^2+10c^2\right)}{\left(5t^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(4b^2+a^2+c^2\right)}+\frac{2}{\left(a^2+b^2+4c^2\right)\left(2t^2+4c^2\right)}\right]\le0\)
Do đó \(f\left(a;b;c\right)\le f\left(t;t;c\right)=f\left(t;t;3-2t\right)\)
\(=\frac{-9\left(t-1\right)^4}{2\left(3t^2-8t+6\right)\left(3t^2-4t+3\right)}+\frac{1}{2}\le\frac{1}{2}\)
Ta có đpcm.
cho a,b,c>0; a+b+c+d=1 chứng minh rằng: \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\le4\sqrt{2}\)
Ta có
\(\sqrt{2}\sqrt{4a+1}\le\frac{4a+3}{2}\)
\(\sqrt{2}\sqrt{4b+1}\le\frac{4b+3}{2}\)
\(\sqrt{2}\sqrt{4c+1}\le\frac{4c+3}{2}\)
\(\sqrt{2}\sqrt{4d+1}\le\frac{4d+3}{2}\)
Cộng vế theo vế ta được
\(\sqrt{2}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\right)\)
\(\le8\)
<=> \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\sqrt{4d+1}\le4\sqrt{2}\)
Bài 1: Cho a,b,c là các số thực dương. Chứng minh rằng:
\(\sqrt{\frac{a+b+4c}{a+b}}+\sqrt{\frac{b+c+4a}{b+c}}+\sqrt{\frac{c+a+4b}{c+a}}\ge3\sqrt{3}.\)
Bài 2:Cho các số thực dương a,b,c thoả mãn abc=1. Chứng minh rằng:
\(\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}+\sqrt[3]{\left(\frac{2b}{bc+1}\right)^2}+\sqrt[3]{\left(\frac{2c}{ca+1}\right)^2}\ge3.\)
Giúp mình với! Mình cần gấp.
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
Bài1 Cho a,b,c >0 và a+b+c = 1
Chứng minh: \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 3\)
Bài 2: Cho x+y = 2 Tìm GTNN của A = \(\frac{1}{x^2+y^2}+\frac{1}{xy}\)
cho a;b;c là các số thực dương thỏa mãn \(a^2+b^2+c^2=\frac{1}{3}\)CMR:\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\ge a+b+c\)