Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Ta có:\(a^5+ab+b^2\ge3a^2b\)
Tương tự ta có:
\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)
\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)
Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)
\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)
Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)
\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
Giả sử \(xy\le1\) thì \(z\ge1\)
Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)
\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)
Dấu = khi \(a=b=c=1\)
Ta có: \(\frac{1}{ab+a+2}=\frac{1}{\left(ab+1\right)+\left(a+1\right)}\)
Áp dụng bất đẳng thức Cauchy-Schwarz dạng cộng mẫu
Ta có: \(\frac{1}{\left(ab+1\right)+\left(a+1\right)}\le\frac{1}{4}\left(\frac{1}{ab+1}+\frac{1}{a+1}\right)\)
\(=\frac{1}{4}\left(\frac{abc}{ab+abc}+\frac{1}{a+1}\right)=\frac{1}{4}\left[\frac{abc}{ab\left(1+c\right)}+\frac{1}{a+1}\right]=\frac{1}{4}\left(\frac{c}{1+c}+\frac{1}{a+1}\right)\) (1)
CMT2 được: \(\frac{1}{bc+b+2}\le\frac{1}{4}\left(\frac{a}{a+1}+\frac{1}{b+1}\right)\) (2)
\(\frac{1}{ca+c+2}\le\frac{1}{4}\left(\frac{b}{b+1}+\frac{1}{c+1}\right)\) (3)
Cộng (1);(2) và (3) vế theo vế
Ta được: \(\frac{1}{ab+a+2}+\frac{1}{bc+b+2}+\frac{1}{ca+c+2}\le\frac{1}{4}\left[\left(\frac{c}{c+1}+\frac{1}{c+1}\right)+\left(\frac{a}{a+1}+\frac{1}{a+1}\right)+\left(\frac{b}{b+1}+\frac{1}{b+1}\right)\right]\)
\(=\frac{1}{4}.\left(1+1+1\right)=\frac{3}{4}\)
=> đpcm
Ta có :
\(a^2+b^2+c^2-2bc-2ca+2ab\)
\(=\left(a+b-c\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2-2bc-2ca+2ab\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge2bc+2ca-2ab\)
Dấu bằng xảy ra khi \(a+b=c\)
Mà \(\frac{5}{3}< \frac{6}{3}=2\)
\(\Rightarrow a^2+b^2+c^2< 2\)
\(\Rightarrow2bc+2ac-2ab\le a^2+b^2+c^2< 2\)
\(\Rightarrow2bc+2ac-2ab< 2\)
Do a ,b , c > 0
\(\Rightarrow\frac{2bc+2ac-2ab}{2abc}< \frac{2}{2abc}\)
\(\Rightarrow\frac{2bc}{2abc}+\frac{2ac}{2abc}-\frac{2ab}{2abc}< \frac{2}{2abc}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)
Vậy ...
Ta có:\(\left(a+b-c\right)^2\ge0\)(với a,b,c > 0)
<=> \(a^2+b^2+c^2+2ab-2bc-2ca\ge0\)
<=> \(bc+ac-ab\le\frac{a^2+b^2+c^2}{2}=\frac{5}{6}< 1\)
Chia 2 vế của bđt cho abc >0 ta dc
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)