Biết a+b+c=2p.CMR:\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}-\dfrac{1}{p}=\dfrac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Có thể giúp mình không ạ!
a) \(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+a\right)\left(1+c\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\) biết abc=1
b) \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c\)
c) \(\dfrac{ab}{a^5+ab+b^5}+\dfrac{bc}{b^5+bc+c^5}+\dfrac{ac}{a^5+ac+c^5}\) biết abc=1
Xin cảm ơn các bạn trước ạ!
b)Ta có: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c\left(1\right)\)
\(\Leftrightarrow\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^4+b^4+c^4}{abc}\ge a+b+c\)
\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Ta xét BĐT phụ: \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
Cộng các BĐT phụ vừa chứng minh:
\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)
Áp dụng vào bài, ta có:
\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)
Áp dụng lần nữa:
\(a^2b^2+b^2c^2+c^2a^2\ge ab^2c+bc^2a+a^2bc=abc\left(a+b+c\right)\)
Vậy ta suy ra được điều phải chứng minh
a) Đặt vế trái BĐT là P
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)8.8}}=\dfrac{3a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(1+a\right)\left(1+c\right)}+\dfrac{1+a}{8}+\dfrac{1+c}{8}\ge\dfrac{3b}{4}\)
\(\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8}\ge\dfrac{3c}{4}\)
Cộng vế theo vế các BĐT vừa chứng minh
\(P+\dfrac{6+2a+2b+2c}{8}\ge\dfrac{3a+3b+3c}{4}\)
\(P\ge\dfrac{3a+3b+3c}{4}-\dfrac{2\left(3+a+b+c\right)}{8}=\dfrac{3a+3b+3c-a-b-c-3}{4}=\dfrac{2\left(a+b+c\right)-3}{4}\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow P\ge\dfrac{2.3-3}{4}=\dfrac{3}{4}\)
Cho a,b,c>0 thỏa abc=1. Chứng minh :
\(\dfrac{a}{\left(a+1\right)^2}+\dfrac{b}{\left(b+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le\dfrac{1}{4}\)
Đành giải tạm bằng nick này vì sợ một vài thành phần trẻ trâu anti phá phách :poor:
Phân tích và giải
Dễ thấy: Dấu "=" khi \(a=b=c=1\)
\(\Rightarrow L=Σ\dfrac{a}{\left(a+1\right)^2}=\dfrac{3}{4}\text{ và }F=-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=-\dfrac{1}{2}\)
Khi đó \(VT=L-F=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)
Ta sẽ chia làm 2 bước cm:
B1: \(Σ\dfrac{a}{\left(a+1\right)^2}\le\dfrac{3}{4}\). Ta xét BĐT :
\(\dfrac{a}{\left(a+1\right)^2}=\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2k}+a^k\right)}{8\left(a^{2k}+a^k+1\right)}\) (cần tìm \(k\) thỏa mãn)
\(\Leftrightarrow8a\left(a^{2k}+a^k+1\right)-3\left(a^{2k}+a^k\right)\left(a^2+2a+1\right)\le0\)\(\Leftrightarrow f\left(a\right)=-3a^{2k}+2a^{k+1}-3a^{k+2}+2a^{2k+1}-3a^{2k+2}-3a^k+8a\)
\(\Rightarrow f'\left(a\right)=2k\cdot-3a^{2k-1}+\left(k+1\right)2a^k-\left(k+2\right)3a^{k+1}+\left(2k+1\right)2a^{2k}-\left(2k+2\right)3a^{2k+1}-k\cdot3a^{k-1}+8a\)
\(\Rightarrow f'\left(1\right)=0\Rightarrow-12k=0\Rightarrow k=0\)
Hay BĐT phụ cần tìm là \(\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2\cdot0}+a^0\right)}{8\left(a^{2\cdot0}+a^0+1\right)}=\dfrac{1}{4}\) (bài này \(k\) đẹp ra luôn \(\farac{1}{4}\) cộng vào là ok =))
\(\Leftrightarrow-\dfrac{\left(a-1\right)^2}{4\left(a+1\right)^2}\le0\) *Đúng* \(\RightarrowΣ\dfrac{a}{\left(a+1\right)^2}\leΣ\dfrac{1}{4}=\dfrac{3}{4}\)
B2: CM \(-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le-\dfrac{1}{2}\)
Tự cm nhé Goodluck :v
Một lời giải sơ cấp:
Đổi \(\left(a;b;c\right)\rightarrow\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\).BDT cần chứng minh tương đương:
\(\sum\dfrac{xy}{\left(x+y\right)^2}-\dfrac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\dfrac{1}{4}\)
\(\Leftrightarrow\left[\dfrac{3}{4}-\sum\dfrac{xy}{\left(x+y\right)^2}\right]+\left[\dfrac{4xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}-\dfrac{1}{2}\right]\ge0\)
\(\Leftrightarrow\sum\left[\dfrac{1}{4}-\dfrac{xy}{\left(x+y\right)^2}\right]-\dfrac{\sum\left(x^2+y^2\right)z-6xyz}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge0\)
\(\Leftrightarrow\sum\dfrac{\left(x-y\right)^2}{4\left(x+y\right)^2}-\dfrac{\sum z\left(x-y\right)^2}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge0\)
\(\Leftrightarrow\sum\left(x-y\right)^2\left[\dfrac{1}{4\left(x+y\right)^2}-\dfrac{z}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\right]\ge0\)
hay \(S_a\left(y-z\right)^2+S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\ge0\)(*)
với \(\left\{{}\begin{matrix}S_a=\dfrac{1}{4\left(y+z\right)^2}-\dfrac{x}{2\prod\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-z\right)}{4\left(y+z\right)^2\left(x+y\right)\left(x+z\right)}\\S_b=\dfrac{1}{4\left(x+z\right)^2}-\dfrac{y}{2\prod\left(x+y\right)}=\dfrac{\left(y-x\right)\left(y-z\right)}{4\left(x+z\right)^2\left(x+y\right)\left(y+z\right)}\\S_c=\dfrac{1}{4\left(x+y\right)^2}-\dfrac{z}{2\prod\left(x+y\right)}=\dfrac{\left(z-x\right)\left(z-y\right)}{4\left(x+y\right)^2\left(y+z\right)\left(z+x\right)}\end{matrix}\right.\)
Dễ thấy \(S_a;S_b;S_c\) không phải là luôn không âm.Giả sử \(x=max\left\{x;y;z\right\}\).
Từ đó suy ra \(S_a\ge0\).Xét \(S_b+S_c=\dfrac{\left(y-z\right)^2}{4\left(x+y\right)^2\left(x+z\right)^2}\ge0,\forall x;y;z>0\)
Do đó \(VT=S_a\left(x-y\right)^2+\left[S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\right]\ge0\)
Ta sẽ chứng minh \(S_b\left(z-x\right)^2+S_c\left(x-y\right)^2\ge0\) với \(S_b+S_c\ge0\)
và điều này đúng hay không e không biết, quan trọng là .. Chúc Mừng Năm Mới !!
a;b;c>0 / abc=1. CMR:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)
Lời giải:
Ta có:
\(\text{VT}=\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\)
\(=\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ac+a+b+c}{abc+(ab+bc+ac)+(a+b+c)+1}\)
\(=\frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\)
Ta cần chứng minh \(\text{VT}\geq \frac{3}{4}\)
\(\Leftrightarrow \frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\geq \frac{3}{4}\)
\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3(ab+bc+ac+a+b+c)+6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\sqrt[6]{ab.bc.ac.a.b.c}\)
(Đúng theo BĐT Cô-si)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=1\)
Thực hiện phép tính:
1) \(A=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
2) \(B=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
3, \(C=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ac}{\left(b-a\right)\left(b-c\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
4) \(D=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Xét:
\(\dfrac{c}{a-b}.\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a^2-b^2\right)}{ab}=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
CMTT cộng theo vế:
\(BTCCM=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}=\dfrac{6\left(a^3+b^3+c^3\right)}{3abc}\)
Mà Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) ( tự cm,ez)
Vậy \(BTCCM=3+6=9\left(đpcm\right)\)
@Nguyễn Thanh Hằng đọc xong xóa đii nha
Cho a,b,c dương.CMR
\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\ge2\left(1+\dfrac{a+b+c}{\sqrt[3]{abc}}\right)\)
Đặt T là vế trái của BĐT, nhân vào biến đổi ta được
\(T=2+\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-3\)
\(T\ge2+\dfrac{2\left(a+b+c\right)}{\sqrt[3]{abc}}+\dfrac{a+b+c}{\sqrt[3]{abc}}-3\)(Sử dụng AM-GM rồi tách)
\(T\ge2+\dfrac{2\left(a+b+c\right)}{\sqrt[3]{abc}}+\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}-3\)
\(T\ge2\left(1+\dfrac{a+b+c}{\sqrt[3]{abc}}\right)\)(đpcm)
Đẳng thức xảy ra khi a=b=c
Cho a,b,c >0 và abc =1. CMR:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)
Giải:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)
\(\Leftrightarrow\) \(\dfrac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)
Do a+1 ; b+1; c+1 >0
\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3
\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0
Áp dụng BĐT Cô-si cho 3 số
Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)
ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3
\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)
\(\Rightarrow\) (*) được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1
Cho \(abc\ne0\) và \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}.\) Tính \(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Giúp ik
Lời giải:
Bạn tham khảo cách làm tương tự tại đây:
https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584
Kết quả $P=8$ hoặc $P=-1$