Cho a,b,c\(\ge\)0 và \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\le1\)
Cmr : \(ab^2c^3\le\frac{1}{5^6}\)
cho a,b,c ≥0 và\(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}\le1\). Chứng minh \(ab^2c^3\le\frac{1}{5^6}\)
\(1-\frac{a}{a+1}\ge\frac{2b}{b+1}+\frac{3c}{c+1}\Leftrightarrow\frac{1}{a+1}\ge\frac{b}{b+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{c}{c+1}+\frac{c}{c+1}\ge5\sqrt[5]{\frac{b^2c^3}{\left(b+1\right)^2\left(c+1\right)^3}}\)
Tương tự:
\(\frac{1}{b+1}\ge\frac{a}{a+1}+\frac{b}{b+1}+3.\frac{c}{c+1}\ge5\sqrt[5]{\frac{abc^3}{\left(a+1\right)\left(b+1\right)\left(c+1\right)^3}}\)
\(\Leftrightarrow\frac{1}{\left(b+1\right)^2}\ge25\sqrt[5]{\frac{a^2b^2c^6}{\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^6}}\)
\(\frac{1}{c+1}\ge\frac{a}{a+1}+2.\frac{b}{b+1}+2.\frac{c}{c+1}\ge5\sqrt[5]{\frac{ab^2c^2}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^2}}\)
\(\Leftrightarrow\frac{1}{\left(c+1\right)^3}\ge125\sqrt[5]{\frac{a^3b^6c^6}{\left(a+1\right)^3\left(b+1\right)^6\left(c+1\right)^6}}\)
Nhân vế với vế:
\(\frac{1}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^3}\ge5^6\sqrt[5]{\frac{a^5b^{10}c^{15}}{\left(a+1\right)^5\left(b+1\right)^{10}\left(c+1\right)^{15}}}=\frac{5^6ab^2c^3}{\left(a+1\right)\left(b+1\right)^2\left(c+1\right)^3}\)
\(\Leftrightarrow ab^2c^3\le\frac{1}{5^6}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{5}\)
Cho a,b,c>0 và \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\le1\)
CMR: \(ab^2c^3d^5\le\frac{1}{10^{11}}\)
Từ \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\le1\)
\(\Rightarrow1-\frac{a}{1+a}+2-\frac{2b}{1+b}+3-\frac{3c}{1+c}+5-\frac{5d}{1+d}\ge10\)
\(\Rightarrow\frac{1}{1+a}+\frac{2}{1+b}+\frac{3}{1+c}+\frac{5}{1+d}\ge10\)
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a+1}\ge\)\(\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\ge10\sqrt[10]{\frac{b^2c^3d^5}{\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}}\)
Và \(\frac{1}{1+b}\ge\)\(\frac{a}{1+a}+\frac{b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\)
\(\ge10\sqrt[10]{\frac{abc^3d^5}{\left(1+a\right)\left(1+b\right)\left(1+c\right)^3\left(1+d\right)^5}}\)
Và \(\frac{1}{1+c}\ge\frac{a}{1+a}+\frac{2b}{b+1}+\frac{2c}{c+1}+\frac{5d}{d+1}\)
\(\ge10\sqrt[10]{\frac{ab^2c^2d^5}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^2\left(1+d\right)^5}}\)
Và \(\frac{1}{1+d}\ge\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{4d}{d+1}\)
\(\ge10\sqrt[10]{\frac{ab^2c^3d^4}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^4}}\)
Nhân theo vế 4 BĐT có: \(\frac{1}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}\)
\(\ge10^{1+2+3+5}\sqrt[10]{\frac{a^{2+3+5}b^{2+2+6+10}c^{3+6+6+15}d^{5+10+15+20}}{\left(1+a\right)^{10}\left(1+b\right)^{20}\left(1+c\right)^{30}\left(1+d\right)^{50}}}\)
Tương đương với \(ab^2c^3d^5\le\frac{1}{10^{11}}\) (ĐPCM)
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Yêu cầu, gợi ý các bạn khác chọn (k) đúng cho mìnhChỉ ghi đáp số mà không có lời giải, hoặc nội dung không liên quan đến câu hỏi.1.\(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=3\end{matrix}\right.\) Cmr: \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{3}{2}\)
2.\(a,b,c>0\). Cmr: \(\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\le\frac{a+b+c}{4}\)
3. \(a,b,c>0\). Cmr: \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{a+b+c}{6}\)
1. Vai trò a, b, c như nhau. Không mất tính tổng quát. Giả sử \(a\ge b\ge0\)
Mà \(ab+bc+ca=3\). Do đó \(ab\ge1\)
Ta cần chứng minh rằng \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(1\right)\)
Và \(\frac{2}{1+ab}+\frac{1}{1+c^2}\ge\frac{3}{2}\left(2\right)\)
Thật vậy: \(\left(1\right)\Leftrightarrow\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\ge0\\ \Leftrightarrow\left(ab-a^2\right)\left(1+b^2\right)+\left(ab-b^2\right)\left(1+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\\ \Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\left(BĐT:đúng\right)\)
\(\left(2\right)\Leftrightarrow c^2+3-ab\ge3abc^2\\ \Leftrightarrow c^2+ca+bc\ge3abc^2\Leftrightarrow a+b+c\ge3abc\)
BĐT đúng, vì \(\left(a+b+c\right)^2>3\left(ab+bc+ca\right)=q\)
và \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\)
Nên \(a+b+c\ge3\ge3abc\)
Từ (1) và (2) ta có \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{2}\)
Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\), ta được
\(\frac{9}{a+3b+2c}=\frac{1}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
Do đó ta được
\(\frac{ab}{a+3b+2c}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)
Hoàn toàn tương tự ta được
\(\frac{bc}{2a+b+3c}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{b+c}+\frac{b}{2}\right);\frac{ac}{3a+2b+c}\le\frac{1}{9}\left(\frac{ac}{a+b}+\frac{ac}{b+c}+\frac{c}{2}\right)\)
Cộng theo vế các BĐT trên ta được
\(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{bc+ab}{a+c}+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}\)Vậy BĐT đc CM
ĐẲng thức xảy ra khi và chỉ khi a = b = c >0
Bài 2:
Áp dụng BĐT AM-GM:
\(a^2+2b^2+c^2=(a^2+b^2)+(a^2+c^2)\geq 2\sqrt{(a^2+b^2)(a^2+c^2)}\geq 2\sqrt{\frac{(a+b)^2}{2}.\frac{(a+c)^2}{2}}=(a+b)(a+c)\)
\(\Rightarrow \frac{ab^2}{a^2+2b^2+c^2}\leq \frac{ab^2}{(a+b)(a+c)}\)
Hoàn toàn tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\leq \sum \frac{ab^2}{(a+b)(a+c)}=\frac{a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)}{(a+b)(b+c)(c+a)}\)
Ta cần CM: \(\frac{a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)}{(a+b)(b+c)(c+a)}\leq \frac{a+b+c}{4}\)
\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)(a+b)(b+c)(c+a)\)
\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)(a+b)(b+c)(c+a)\)
\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)[(a+b+c)(ab+bc+ac)-abc]\)
\(\Leftrightarrow 2(a^2b^2+b^2c^2+c^2a^2)\leq (a^3b+ab^3)+(bc^3+b^3c)+(ca^3+c^3a)\)
(dễ thấy luôn đúng do theo BĐT AM-GM)
Do đó ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Cho a,b,c>0 CMR:\(\frac{a}{3a^2+2b^2+c^2}+\frac{b}{3b^2+2c^2+a^2}+\frac{c}{3c^2+2a^2+b^2}\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
cho a,b,c >0
CMR:\(\frac{a}{3a^2+2b^2+c^2}+\frac{b}{3b^2+2c^2+a^2}+\frac{c}{3c^2+2a^2+b^2}\le\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cho a, b, c > 0:
CMR: \(\frac{1}{5a+b}+\frac{1}{5b+c}+\frac{1}{5c+a}\ge\frac{1}{a+3b+2c}+\frac{1}{b+3c+2a}+\frac{1}{c+3a+2b}\)
Xin ngoại lệ ạ ( Ko liên quan đến câu hỏi)
Cho a,b,c là các số dương thỏa mãn: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=6\). CMR:
a) \(\frac{1}{a+b+2c}+\frac{1}{b+c+2a}+\frac{1}{c+a+2b}\le3\)
b) \(\frac{1}{3a+3b+2c}+\frac{1}{3a+2b+3c}+\frac{1}{2a+3b+2c}\le\frac{3}{2}\)
Ta CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},a+b\ge2\sqrt{ab}\)( co si với a,b>0)
Suy ra \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(a+b\right)\ge4\RightarrowĐPCM\)\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)
a/Áp dụng (1) có
\(\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\left(2\right)\).Tương tự ta cũng có:
\(\frac{1}{b+c+2a}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\left(3\right),\frac{1}{c+a+2b}\le\frac{1}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\left(4\right)\)
Cộng (2),(3) và (4) có \(VT\le\frac{1}{4}.\left(6+6\right)=3\left(ĐPCM\right)\)
b/Áp dụng (1) có:
\(\frac{1}{3a+3b+2c}=\frac{1}{\left(a+b+2c\right)+2\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{2\left(a+b\right)}\right)\left(5\right)\)
Tương tự có: \(\frac{1}{3a+2b+3c}\le\frac{1}{4}\left(\frac{1}{a+c+2b}+\frac{1}{2\left(a+c\right)}\right)\left(6\right)\)
\(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{2a+b+c}+\frac{1}{2\left(b+c\right)}\right)\left(7\right)\)
Cộng (5),(6) và (7) có:
\(VT\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{a+c+2b}+\frac{1}{2a+b+c}+\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\right)\le\frac{1}{4}.9=\frac{3}{2}\)
1. Cho a,b,c > 0. Cmr :
\(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
2. Cho a,b,c > 0. Cmr :
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
Thục Trinh, tran nguyen bao quan, Phùng Tuệ Minh, Ribi Nkok Ngok, Lê Nguyễn Ngọc Nhi, Tạ Thị Diễm Quỳnh,
Nguyễn Huy Thắng, ?Amanda?, saint suppapong udomkaewkanjana
Help me!
Bài thứ hai đó áp dụng bđt cauchy showas là ra rồi sử dụng tch bắc cầu tệ.
Cho a,b,c >0 thoar manx ab + bc + ca =5abc
CMR: \(P=\frac{1}{2a+2b+2c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\le1\)
Từ \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Áp dụng BĐT Bu-nhi-a-cốp-xki ta có :
\(\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\left(a+a+b+b+c\right)\ge\left(1+1+1+1+1\right)^2\)
\(\Rightarrow\frac{2}{a}+\frac{2}{b}+\frac{1}{c}\ge\frac{25}{2a+2b+c}\)
Tương tự ta có :
\(\frac{2}{b}+\frac{2}{c}+\frac{1}{a}\ge\frac{25}{2b+2c+a}\)
\(\frac{2}{a}+\frac{1}{b}+\frac{2}{c}\ge\frac{25}{2a+b+2c}\)
Cộng từng vế BĐT ta thu được :
\(\frac{5}{a}+\frac{5}{b}+\frac{5}{c}\ge25P\)
\(\Leftrightarrow P\le\frac{5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{25}=1\)
Vậy BĐT đã được chứng minh . Dấu \("="\) xảy ra khi \(a=b=c=\frac{3}{5}\)