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Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:
\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)
\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)
\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)
Cộng (1),(2) và (3) có:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Áp dụng BĐT Bunhiacopxki ta có :
\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có : \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\Leftrightarrow a+b+c\ge3\) ( tự chứng minh ạ )
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Áp dụng BĐT Cachy Schwarz ta có :
\(\frac{a^4}{b+3c}+\frac{b^4}{c+3a}+\frac{c^4}{a+3b}\ge\frac{\left(a^2+b^2+c^2\right)^2}{4\left(a+b+c\right)}\) \(\ge\frac{\left[\frac{\left(a+b+c\right)}{3}\right]^2}{4\left(a+b+c\right)}=\frac{\left(a+b+c\right)^3}{36}\)
\(\ge\frac{27}{36}=\frac{3}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\) ( bạn tự giải rõ ạ )
1.
Áp dụng bất đẳng thức Cô-si thôi:
\(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}\ge\frac{2\sqrt{ab}}{ab}=\frac{2}{\sqrt{ab}}\ge\frac{2}{\frac{a+b}{2}}=\frac{4}{a+b}\)
Dấu "=" khi a = b
2.
Vì a,b,c là ba cạnh tam giác nên dễ thấy các mẫu số dương.
Áp dụng câu 1 ta có:
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{4}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)
Tương tự:
\(\frac{1}{c+a-b}+\frac{1}{b+c-a}\ge\frac{4}{2c}=\frac{2}{c}\)
\(\frac{1}{b+c-a}+\frac{1}{a+b-c}\ge\frac{4}{2b}=\frac{2}{b}\)
Cộng theo vế ta được:
\(2\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)
Dấu "=" xảy ra khi a = b = c hay tam giác đó đều.
#)Giải :
Ta có :
\(\hept{\begin{cases}\frac{ab}{b+c+a+b}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\\frac{bc}{a+b+a+c}\le\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{ac}{b+c+a+b}\le\frac{ac}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{a+b}.\left(\frac{bc}{4}+\frac{ac}{4}\right)+\frac{1}{a+c}.\left(\frac{bc}{4}+\frac{ab}{4}\right)+\frac{1}{b+c}.\left(\frac{ac}{4}+\frac{ab}{4}\right)\)
\(=\frac{1}{a+b}.\frac{c\left(a+b\right)}{4}+\frac{1}{a+c}.\frac{b\left(a+c\right)}{4}+\frac{1}{b+c}.\frac{a\left(b+c\right)}{4}\)
\(=\frac{c}{4}+\frac{b}{4}+\frac{a}{4}\)
\(\Rightarrow\frac{a+b+c}{4}\)
\(\Rightarrowđpcm\)
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
a, Ta cần phải chứng minh (a+b)(\(\frac{1}{a}+\frac{1}{b}\))=1+\(\frac{a}{b}+\frac{b}{a}+1=2+\frac{a}{b}+\frac{b}{a}\ge4\) vì
\(\frac{a}{b}+\frac{b}{a}\ge2\)(cái này bạn tìm hiểu kĩ hơn nha,nhưng mk nghĩ thế này đc rồi đó)
Dấu ''='' xảy ra \(\Leftrightarrow\)a=b.
d,(a+b+c)(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\))=1+\(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)
=3+(\(\frac{a}{b}+\frac{b}{a}\))+(\(\frac{a}{c}+\frac{c}{a}\))+(\(\frac{c}{b}+\frac{b}{c}\))\(\ge\)3+2+2+2=9
Dấu ''='' xảy ra \(\Leftrightarrow\)a=b=c
e,Xét hiệu :
\(^{a^3+b^3+c^3-3abc=\left(a^2+b^2+c^2-ab-ac-bc\right)\left(a+b+c\right)}\) => cái này bạn nhân ra trước rồi phân tích đa thức thành nhân tử nha.
=\(\left(a+b+c\right)\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\ge0\) \(\Rightarrow\)ĐPCM
Câu b) khó hiểu, ko có dấu "=" nhaaaa !!!!!!
\(B=3\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{a}{b}+\frac{b}{a+b}\ge3.2+\frac{a}{b}+\frac{b}{a+b}=6+\frac{a}{b}+\frac{b}{a+b}>6+0+0=6\)
Vậy ta thấy \(B>6\)
=> Ko có dấu "=" xảy ra.