Cho a,b,c>0 thỏa mãn abc=1. Chứng minh:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2}{3}\left[\dfrac{1}{a^3bc\left(b^2+1\right)}+\dfrac{1}{b^3ca\left(c^2+1\right)}+\dfrac{1}{c^3ab\left(a^2+1\right)}\right]\).
Cho a, b, c > 0 và abc = 1. Chứng minh rằng \(\dfrac{1}{a^2.\left(b+c\right)}+\dfrac{1}{b^2.\left(c+a\right)}+\dfrac{1}{c^2.\left(a+b\right)}\ge\dfrac{3}{2}\)
Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)
Ta có:
\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)
BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)
Đánh giá cuối cùng đúng theo BĐT Cauchy
Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
\(\text{Cho a,b,c >0 thỏa a+b+c=1.Chứng minh:}\)
\(\Sigma_{cyc}\dfrac{a}{1+3bc+4\left(b+c\right)}\ge\dfrac{1}{2}\)
\(VT=\sum\dfrac{a^2}{a+3abc+4\left(ab+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+9abc+8\left(ab+bc+ca\right)}\)
\(VT\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\dfrac{1}{3}\left(a+b+c\right)^3+\dfrac{8}{3}\left(a+b+c\right)^2}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\dfrac{1}{3}\)
Đề bài bị sai con số bên vế phải
Cho 3 số thực dương x,y,z thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\)
Chứng minh \(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^3}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng: \(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\dfrac{1}{2}\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
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 !!
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Cho a,b,c thỏa mãn ab+bc+ca =1. Chứng minh rằng
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Ta có VP:
\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Thay \(1=ab+bc+ca\)
\(=\dfrac{2}{\sqrt{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}}\)
\(=\dfrac{2}{\sqrt{\left[b\left(a+c\right)+a\left(a+c\right)\right]\left[a\left(b+c\right)+b\left(b+c\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]}}\)
\(=\dfrac{2}{\sqrt{\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(a+c\right)}}\)
\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)
\(=\dfrac{2}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)
_____________
Ta có VT:
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)
Thay \(1=ab+ac+bc\)
\(=\dfrac{a}{ab+ac+bc+a^2}+\dfrac{b}{ab+ac+bc+b^2}+\dfrac{c}{ab+ac+bc+c^2}\)
\(=\dfrac{a}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b}{b\left(b+c\right)+a\left(b+c\right)}+\dfrac{c}{c\left(b+c\right)+a\left(b+c\right)}\)
\(=\dfrac{a}{\left(a+c\right)\left(a+b\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{a\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}+\dfrac{b\left(a+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\dfrac{c\left(a+b\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{ab+ac+ab+bc+ac+bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2ab+2ac+2bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2\cdot\left(ab+ac+bc\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\left(ab+ac+bc=1\right)\)
Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\left(dpcm\right)\)
Cho a, b, c là ba số dương thỏa mãn \(abc\)=1. Chứng minh rằng:
\(\dfrac{1}{a^3\left(b+c\right)}\)+\(\dfrac{1}{b^3\left(a+c\right)}\)+\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
Cho các số dương a,b,c cs abc=1 Chứng minh rằng
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}\ge\dfrac{1}{4}\)
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)