Cho biết a + b + c = 2p
CMR: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
cho \(a+b+c=2p.cmr\frac{1}{\left(p-a\right)}+\frac{1}{\left(p-b\right)}+\frac{1}{\left(p-c\right)}+\frac{1}{p}=\frac{abc}{\left(p-a\right)\left(p-b\right)\left(p-c\right)p}\)
Sửa đề: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
Ta có:
\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{2p^3-\left(a+b+c\right)p^2+abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
\(=\frac{2p^3-2p.p^2+abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
cho a,b,c dương thỏa abc=1
chứng minh \(\frac{a}{\left(a+1\right)^2}+\frac{b}{\left(b+1\right)^2}+\frac{c}{\left(c+1\right)^2}-\frac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le\frac{1}{4}\)
Đặt: \(\hept{\begin{cases}\frac{1-a}{1+a}=x\\\frac{1-b}{1+b}=y\\\frac{1-c}{1+c}=z\end{cases}}\)
\(\Rightarrow-1< x,y,z< 1\)và \(\hept{\begin{cases}\frac{1-x}{1+x}=a\\\frac{1-y}{1+y}=b\\\frac{1-z}{1+z}=c\end{cases}}\)
Theo đề bài ta có: \(abc=1\Rightarrow\left(1-x\right)\left(1-y\right)\left(1-z\right)=\left(1+x\right)\left(1+y\right)\left(1+z\right)\)
\(\Rightarrow x+y+z+xyz=0\)
Mặt khác ta có: \(\frac{4a}{\left(a+1\right)^2}=1-x^2;\frac{2}{a+1}=1+x\)
Và: \(\frac{4b}{\left(b+1\right)^2}=1-y^2;\frac{2}{b+1}=1+y\)
Và: \(\frac{4c}{\left(c+1\right)^2}=1-z^2;\frac{2}{c+1}=1+z\)
Nên: \(\frac{4a}{\left(a+1\right)^2}+\frac{4b}{\left(b+1\right)^2}+\frac{4c}{\left(c+1\right)^2}\le1+2.\frac{2}{a+1}.\frac{2}{b+1}.\frac{2}{c+1}\)
\(\Leftrightarrow x^2+y^2+z^2+\left(xy+yz+zx\right)+2\left(x+y+z+xyz\right)\ge0\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge0\)
Đây là BĐT luôn đúng nên ta có đpcm.
ミ★ᗪเệų ℌųуềй (ßăйǥ ßăйǥ ²к⁶)★彡 Giải ghê quá, t chẳng hiểu gì.
Đặt \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\)
BĐT \(\Leftrightarrow \sum\limits_{cyc} \frac{xy}{(x+y)^2} \leq \frac{1}{4}+\frac{4xyz}{(x+y)(y+z)(z+x)}\)
Ta có: \(VP-VT=\frac{4\left(x-y\right)^2\left(y-z\right)^2\left(z-x\right)^2}{4\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\ge0\)
BĐT hiển nhiên đúng.
Ôi trời, dòng 3 gõ latex mà olm không hiện à?
BĐT \(\Leftrightarrow\Sigma_{cyc}\frac{xy}{\left(x+y\right)^2}-\frac{1}{4}\le\frac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng:
\(\frac{a^3}{\left(b+1\right)\left(c+2\right)}+\frac{b^3}{\left(c+1\right)\left(a+2\right)}+\frac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\frac{1}{2}\)
Ta có \(a+b+c\ge3\sqrt[3]{abc}=3\)
Áp dụng bđt cosi ta có:
\(\frac{a^3}{\left(b+1\right)\left(c+2\right)}+\frac{b+1}{12}+\frac{c+2}{18}\ge3\sqrt[3]{\frac{a^3}{12.18}}=\frac{a}{2}\)
Làm tương tự
=>\(VT+\left(\frac{a+1}{12}+\frac{a+2}{18}\right)+\left(\frac{b+1}{12}+\frac{b+2}{18}\right)+\left(\frac{c+1}{12}+\frac{c+2}{18}\right)\ge\frac{a+b+c}{2}\)
=> \(VT\ge\frac{13}{36}.\left(a+b+c\right)-\frac{7}{12}\ge\frac{13}{36}.3-\frac{7}{12}=\frac{1}{2}\)(ĐPCM)
Cho biết a+b+c=2p
CMR :\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}+\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
cho a, b, c dương. chứng minh
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(c+a\right)}\ge\frac{3}{\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
Cho biết a+b+c =2p
CMR \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}-\frac{1}{p}=\frac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)c)
các bạn làm được ý nào thì làm ý đó nha
1. Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh:
a) \(\frac{1}{\left(a+b-c\right)^2}+\frac{1}{\left(a-b+c\right)^2}+\frac{1}{\left(b+c-a\right)^2}\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
b) \(\frac{1}{\left(a+b-c\right)^3}+\frac{1}{\left(a-b+c\right)^3}+\frac{1}{\left(b+c-a\right)^3}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
c) \(\frac{1}{\left(a+b-c\right)^{200}}+\frac{1}{\left(a-b+c\right)^{200}}+\frac{1}{\left(b+c-a\right)^{200}}\ge\frac{1}{a^{200}}+\frac{1}{b^{200}}+\frac{1}{c^{200}}\)
d) \(\frac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}\)
e) \(a+b+c< \sqrt{a\left(b+c\right)}+\sqrt{b\left(a+c\right)}+\sqrt{c\left(a+b\right)}\)
f) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}< \sqrt{6}\)
g) \(\sqrt{-a+b+c}+\sqrt{a-b+c}+\sqrt{a+b-c}\le\sqrt{3\left(a+b+c\right)}\)
cho a,b,c>0
Cm: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)
\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
Cho a,b,c dương thõa mãn abc=1
CMR \(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)
Ta có:
\(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{a\left(a+1\right)}{8}+\frac{a\left(b+1\right)}{8}\ge3\sqrt[3]{\frac{a^3\left(a+1\right)\left(b+1\right)}{64\left(a+1\right)\left(b+1\right)}}=\frac{3a}{4}\)
\(\Rightarrow LHS+\frac{a^2+b^2+c^2+ab+bc+ca+2\left(a+b+c\right)}{8}\ge\frac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow LHS\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(a+b+c\right)-\frac{a^2+b^2+c^2+ab+bc+ca}{8}\)
\(\ge\frac{a+b+c}{2}-\frac{a^2+b^2+c^2}{4}\)
Có ý tưởng đến đây thôi nhưng lại bị ngược dấu rồi :(
BĐT <=> \(\frac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{3}{4}\)
<=> \(\frac{ab+bc+ac+a+b+c}{abc+1+ab+bc+ac+a+c+b}\ge\frac{3}{4}\)
<=> \(4\left(ab+bc+ac+a+b+c\right)\ge3\left(ab+bc+ac+a+b+c+2\right)\)
<=> \(ab+bc+ac+a+b+c\ge6\)(1)
(1) luôn đúng do \(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3;a+b+c\ge3\sqrt[3]{abc}=3\)
=> BĐT được CM
Dấu bằng xảy ra khi \(a=b=c=1\)