Cho a,b,c,d > 0.
Cmr: \(\dfrac{a^3}{a^3+3bcd}+\dfrac{b^3}{b^3+3cda}+\dfrac{c^3}{c^3+3dab}+\dfrac{d^3}{d^3+3abc}\ge1\)
\(\text{cho a,b,c,d thỏa mãn }\)\(c^2+d^2=(a^2+b^2)^3\).CMR \(\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge1\)
Mn giúp e với
BĐT mà ghi thiếu điều kiện thì chết rồi, vì số thực, số dương, số không âm nó hoạt động khác nhau lắm
Bunhiacopxki: \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(a^2+b^2\right)^4\)
\(\Rightarrow ac+bd\le\left(a^2+b^2\right)^2\)
Do đó:
\(\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^4}{ac}+\dfrac{b^4}{bd}\ge\dfrac{\left(a^2+b^2\right)^2}{ac+bd}\ge\dfrac{\left(a^2+b^2\right)^2}{\left(a^2+b^2\right)^2}=1\) (đpcm)
Đề bài sai: phản ví dụ:
\(a=b=-1\) ; \(c=d=2\)
Khi đó: \(c^2+d^2=\left(a^2+b^2\right)^3\) nhưng \(\dfrac{a^3}{c}+\dfrac{b^3}{d}=-1< 1\)
cho a,b,c>0 Sao cho a+b+c=3
CMR \(\dfrac{a^3}{a+2b^3}+\dfrac{b^3}{b+2c^3}+\dfrac{c^3}{c+2a^3}\ge1\)
Đặt A=\(\sum\dfrac{a^3}{a+2b^3}\)
Ta có \(a^3+1+1\ge3a\Rightarrow a\le\dfrac{a^3+2}{3}\)\(\Rightarrow\sum\dfrac{a^3}{a+2b^3}\ge\sum\dfrac{a^3}{\dfrac{a^3+2}{3}+2b^3}=\sum\dfrac{3a^3}{a^3+6b^3+2}\)
Đặt \(a^3=x;b^3=y;c^3=z,taco:x+y+z\ge3\)
Mà A=\(3\left(\sum\dfrac{x}{x+6y+2}\right)=3\left(\sum\dfrac{x^2}{x^2+6xy+2x}\right)\ge3\dfrac{\left(x+y+z\right)^2}{\sum x^2+\sum6xy+2\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+4\left(xy+yz+zx\right)+2\left(x+y+z\right)}\)
Mà \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\), đặt x+y+z=m
Ta có \(A\ge\dfrac{3m^2}{m^2+\dfrac{4}{3}m^2+m}\), cần \(\dfrac{3m^2}{\dfrac{7}{3}m^2+2m}\ge1\Leftrightarrow3m^2\ge\dfrac{7}{3}m^2+2m\Leftrightarrow\dfrac{2}{3}m\ge2\Leftrightarrow m\ge1\left(LĐ\right)\)
=> BDT cần chứng minh luôn đúng
dấu = xảy ra <=> a=b=c=1
Bài 1
CMR: \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}cmr:\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\)
Lời giải:
Vì \(\frac{a}{c}=\frac{c}{b}=\frac{b}{d}\)
\(\Rightarrow \frac{a^3}{c^3}=\frac{c^3}{b^3}=\frac{b^3}{d^3}=\frac{a^3+c^3-b^3}{c^3+b^3-d^3}(1)\) (theo tính chất dãy tỉ số bằng nhau)
Mặt khác:
\(\frac{a}{c}=\frac{c}{b}=\frac{b}{d}\Rightarrow \frac{a}{c}.\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{c}{b}.\frac{b}{d}\)
Hay \(\frac{a^3}{c^3}=\frac{a}{d}(2)\)
Từ \((1);(2)\Rightarrow \frac{a^3+c^3-b^3}{c^3+b^3-d^3}=\frac{a}{d}\) (đpcm)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}\)
CMR:\(\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}=\dfrac{a^3}{c^3}=\dfrac{c^3}{b^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}\left(1\right)\)
Từ \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}\)
Ta xét tích: \(\left(\dfrac{a}{c}\right)^3=\dfrac{a}{c}.\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\left(dpcm\right)\)
a,\(Cho\dfrac{a}{b}=\dfrac{c}{d}CMR,\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(2)
Từ (1) và (2) \(\Rightarrow\) đpcm
Theo đề đã cho, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)(1)
\(\Rightarrow\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(2)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(đpcm)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{\left[b\left(k+1\right)\right]^3}{\left[d\left(k+1\right)\right]^3}=\dfrac{b^3}{d^3}\\\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{bk^3+b^3}{dk^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\end{matrix}\right.\)
Vậy
Cho b^2 = ac ; c^2 = bd với b, c, d ≠ 0; b+c ≠ 0; b^3+c^3≠ d^3 3. Chứng minh rằng:
a) \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
b) \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Cho a, b, c>0; abc=1. Cmr:
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\ge1\)
Sao em làm chỉ ra >=3 thôi ạ)):
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3b\left(b+2\right)}{27b\left(c+2\right)}}=a\)
Tương tự: \(\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c}{3}+\dfrac{a+2}{9}\ge b\)
\(\dfrac{c^3}{a\left(b+2\right)}+\dfrac{a}{3}+\dfrac{b+2}{9}\ge c\)
Cộng vế:
\(VT+\dfrac{4\left(a+b+c\right)}{9}+\dfrac{2}{3}\ge a+b+c\)
\(\Rightarrow VT\ge\dfrac{5\left(a+b+c\right)}{9}-\dfrac{2}{3}\ge\dfrac{15}{9}-\dfrac{2}{3}=1\)
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
Cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}\)
CMR \(\dfrac{a^3+c^3+m^3}{b^3-d^3-n^3}\) = \(\left(\dfrac{a+c-m}{b+d-m}\right)^3\)
mọi người ơi giup mik với ai làm đc mik tick cho
Sửa: CMR \(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}=k\Rightarrow a=kb;c=kd;m=kn\)
\(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\dfrac{k^3b^3+k^3d^3+k^3n^3}{b^3+d^3+n^3}=\dfrac{k^3\left(b^3+d^3+n^3\right)}{b^3+d^3+n^3}=k^3\)
\(\left(\dfrac{a+c-m}{b+d-m}\right)^3=\left(\dfrac{kb+kd-kn}{b+d-n}\right)^3=\left(\dfrac{k\left(b+d-n\right)}{b+d-n}\right)^3=k^3\)
\(\Rightarrow\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\left(=k^3\right)\)