Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lần sau đăng ít một thôi toàn bài dài :v, ko phải ko làm mà là ngại làm
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{2a+b+c}=\frac{a}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{a+2b+c}\le\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right);\frac{c}{a+b+2c}\le\frac{1}{4}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{4}\)
Xảy ra khi \(a=b=c\)
b)Đặt \(THANG=abc\left(a^2+bc\right)\left(b^2+ac\right)\left(c^2+ab\right)>0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b+c}{a^2+bc}-\frac{c+a}{b^2+ac}-\frac{a+b}{a^2+ab}\)
\(=\frac{a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-b^4c^2a^2-c^4a^2b^2}{THANG}\)
\(=\frac{\left(a^2b^2-b^2c^2\right)^2+\left(b^2c^2-c^2a^2\right)+\left(c^2a^2-a^2b^2\right)^2}{2THANG}\ge0\) (Đúng)
Xảy ra khi \(a=b=c\)
c)Ta có:\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)
Và \(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ab\left(b-a\right)}{\left(c+a\right)\left(c^2+a^2\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(b+a\right)\left(b^2+a^2\right)}\)
Cộng theo vế 3 đăng thức trên ta có:
\(VT-VP=Σ\left[\frac{ab\left(a-b\right)}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{ab\left(a-b\right)}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)
\(=\left(a^2+b^2+c^2+ab+bc+ca\right)\cdotΣ\frac{ab\left(a-b\right)^2}{\left(b+c\right)\left(c+a\right)\left(b^2+c^2\right)\left(c^2+a^2\right)}\ge0\)
2 bài cuối full quy đồng mệt thật :v
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\ge2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)
\(P=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\)
\(\Rightarrow P\ge\sqrt{\frac{\left(x+2y\right)^2}{3}}+\sqrt{\frac{\left(y+2z\right)^2}{3}}+\sqrt{\frac{\left(z+2x\right)^2}{3}}\)
\(\Rightarrow P\ge\frac{1}{\sqrt{3}}\left(3x+3y+3z\right)\ge\frac{3}{\sqrt{3}}=\sqrt{3}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
Bạn tham khảo:
Câu hỏi của Phạm Vũ Trí Dũng - Toán lớp 8 | Học trực tuyến
#)Giải :
Ta có :
\(\hept{\begin{cases}\frac{ab}{b+c+a+b}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\\frac{bc}{a+b+a+c}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{ac}{b+c+a+b}\le\frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{a+b}.\left(\frac{bc}{4}+\frac{ac}{4}\right)+\frac{1}{a+c}.\left(\frac{bc}{4}+\frac{ab}{4}\right)+\frac{1}{b+c}.\left(\frac{ac}{4}+\frac{ab}{4}\right)\)
\(=\frac{1}{a+b}.\frac{c\left(a+b\right)}{4}+\frac{1}{a+c}.\frac{b\left(a+c\right)}{4}+\frac{1}{b+c}.\frac{a\left(b+c\right)}{4}\)
\(=\frac{c}{4}+\frac{b}{4}+\frac{a}{4}\)
\(\Rightarrow\frac{a+b+c}{4}\)
\(\Rightarrowđpcm\)
\(\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\left(1\right)\)
Ta có ab+bc+ca=abc nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
\(\left(1\right)\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\ge\sqrt{3}\)
Trong mặt phẳng với hệ tọa độ Oxy, với các Vecto
\(\overrightarrow{u}=\left(\frac{1}{a};\frac{\sqrt{2}}{b}\right);\left|\overrightarrow{u}\right|=\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}\)
\(\overrightarrow{v}=\left(\frac{1}{b};\frac{\sqrt{2}}{c}\right)\Rightarrow\left|\overrightarrow{v}\right|=\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}\)
\(\overrightarrow{w}=\left(\frac{1}{c};\frac{\sqrt{2}}{a}\right)\Rightarrow\left|\overrightarrow{w}\right|=\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\)
Ta có \(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c};2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)=\left(1;\sqrt{2}\right)\)
=> \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|=\sqrt{1+2}=\sqrt{3}\)
Mặt khác \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\ge\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\)
\(\Rightarrow\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ac}\ge\sqrt{3}\)
Dấu "=" xảy ra <=> a=b=c
Cho a,b,c>0.Chung minh rang \(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}\)
Ta có:
\(\left(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\right)\left[\left(b+2c\right)+\left(c+2a\right)+\left(a+2b\right)\right]\)
\(\ge\left[\sqrt{\frac{a^2}{b+2c}.\left(b+2\right)}+\sqrt{\frac{b^2}{c+2a}.\left(c+2a\right)}+\sqrt{\frac{c^2}{a+2b}.\left(a+2b\right)}\right]^2\)
\(=\left(a+b+c\right)^2\)
\(\Rightarrow\left(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\right)\left[3\left(a+b+c\right)\right]\ge\left(a+b+c\right)^2\)
\(\Rightarrow\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}\left(đpcm\right)\)
Áp dụng bđt Cauchy-Schwarz:
\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\ge\frac{\left(1+1+1\right)^2}{2a+b+c+a+2b+c+a+b+2c}=\frac{9}{4a+4b+4c}\)Dấu "=" xảy ra khi a=b=c