Cho cac so a,b,c,d duong va a.b.c.d=1 .CM
\(\left(a+b\right).\left(b+c\right).\left(c+d\right)\)\(.\left(d+a\right)\ge16\)
Cho a,b,c,d là số dương. Cmr
a/ \(\left(\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{d^3}+\dfrac{d}{a^3}\right)\left(a+b\right)\left(b+c\right)\ge16\)
b/ \(\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
a) sai đề
b) để ý rằng :Theo AM-GM
\(VT=\dfrac{a+b}{2\sqrt[3]{abc}}+\dfrac{b+c}{2\sqrt[3]{abc}}+\dfrac{c+a}{2\sqrt[3]{abc}}+\dfrac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge4\)
Dấu = xảy ra khi a=b=c.
P/s: Min ra xấp xỉ \(14,4809\)( wolframalpha.com)
cho cac so thuc duong a b c thoa a^2+b^2+c^2>=3 chung minh
\(\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}+\frac{\left(b+1\right)\left(c+2\right)}{\left(c+1\right)\left(c+5\right)}+\frac{\left(c+1\right)\left(a+2\right)}{\left(a+1\right)\left(a+5\right)}\ge\frac{3}{2}\)
Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)
Thật vậy, BĐT trên tương đương:
\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)
\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)
\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)
Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)
\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)
\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)
\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c la cac so duong thoa man dieu kien
a+b+c=1
Tim GTNN :
A=\(\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)
Rút gọn biểu thức sau:
\(\frac{\left(a.b+b.c+c.d+d.a\right).a.b.c.d}{\left(c+d\right).\left(a+b\right)+\left(b-c\right).\left(a-d\right)}\)
Rút gọn biểu thức sau: \(\frac{\left(a.b+b.c+c.d+d.a\right).a.b.c.d}{\left(c+d\right).\left(a+b\right)+\left(b-c\right).\left(a-d\right)}\)
Nhân mẫu số vào ta được :
ac + ad + bd + bc +ab - ac -bd + dc = ab + bc + cd +da
=> biểu thức trên có giá trị rút gọn là abcd
Cho \(\frac{a}{b}\)= \(\frac{b}{c}\)= \(\frac{c}{d}\)= \(\frac{d}{a}\). Tính M = \(\frac{\left(a+b\right)\left(b+c\right)\left(c+d\right)\left(d+a\right)}{a.b.c.d}\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{a+b+c+d}=1\)
\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=d\end{cases}}\Rightarrow a=b=c=d\)
\(\Rightarrow M=\frac{\left(2a\right)^4}{a^4}=16\)
Cho a.b.c.d \(\in\)R. CMR:
a) \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge8abc\)
b) \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge256abcd\)
a) Ta có: \(a^2+1\ge2a\)
Tường tự \(b^2+1\ge2b\); \(c^2+1\ge2c\)
Vì \(a^2+1\ge0\);\(b^2+1\ge0\);\(c^2+1\ge0\)nên ta:
Nhân vế theo vế của 3 bất đẳng thức cùng chiều ta được điều phải chứng minh
b) \(a^2+2^2\ge4a\)bạn làm tương tự như câu a) là ra nha!
Cho 5 số thực khác nhau a,b,c,d,x.Chứng minh :
\(\frac{b+c+d}{\left(b-a\right)\left(c-a\right)\left(d-a\right)\left(x-a\right)}+\frac{a+c+d}{\left(a-b\right)\left(c-b\right)\left(d-b\right)\left(x-b\right)}+\frac{a+b+d}{\left(a-c\right)\left(b-c\right)\left(d-c\right)\left(x-c\right)}+\)
\(\frac{a+b+c}{\left(a-d\right)\left(b-d\right)\left(c-d\right)\left(x-d\right)}=\frac{a+b+c+d-x}{\left(a-x\right)\left(b-x\right)\left(c-x\right)\left(d-x\right)}\)
cho a,b,c,d là các số dương . CMR :
\(\frac{abc}{\left(a+d\right)\left(b+d\right)\left(c+d\right)}+\frac{bcd}{\left(b+a\right)\left(c+a\right)\left(d+a\right)}+\frac{cda}{\left(a+b\right)\left(c+b\right)\left(d+b\right)}+\frac{dab}{\left(d+c\right)\left(a+c\right)\left(b+c\right)}\ge\frac{1}{2}\)