\(a,b,c\ge1.CMR:\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
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Cho a,b,c\(\ge1\)CMR \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{1}{1+\sqrt[4]{ab^3}}+\frac{1}{1+\sqrt[4]{bc^3}}+\frac{1}{1+\sqrt[4]{ca^3}}\)
Ta có \(a+b+b+b\ge4\sqrt[4]{abbb}\)(theo BĐT Cosi)
\(\Leftrightarrow a+3b\ge\sqrt[4]{ab^3}\)
\(\Leftrightarrow\frac{a+3b}{4}\ge4\sqrt[4]{ab^3}\)
Mà \(a,b,c\ge1\Rightarrow a+3b\ge4\Rightarrow\frac{a+3b}{4}\ge1\)
\(\Leftrightarrow1+\sqrt[4]{ab^3}\ge1+a\)
\(\Rightarrow\frac{1}{1+\sqrt[4]{ab^3}}\le\frac{1}{1+a}\left(1\right)\)
Tương tự \(\hept{\begin{cases}\frac{1}{1+\sqrt[4]{bc^3}}=\frac{1}{1+b}\left(2\right)\\\frac{1}{1+\sqrt[4]{ca^3}}=\frac{1}{1+c}\left(3\right)\end{cases}}\)
(1) (2) (3) => \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge\frac{1}{1+\sqrt[4]{ab^3+1}}+\frac{1}{1+\sqrt[4]{bc^3}}+\frac{1}{1+\sqrt[4]{ca^3}}\)(đpcm)
cho a,b>0 cm\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\) nếu \(ab\ge1\)
b) cho a,b,c\(\ge\)1. CMR \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)
\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)
b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)
\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)
Cộng vế với vế ta có đpcm
cho a, b, c ≥ 1
cmr: \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Sử dụng BĐT quen thuộc: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\) với \(xy\ge1\)
\(2VT\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^2c^2}+\frac{2}{1+c^2a^2}\)
\(\Rightarrow VT\ge\frac{1}{1+a^2b^2}+\frac{1}{1+b^2c^2}+\frac{1}{1+c^2a^2}\)
\(\Rightarrow2VT\ge\frac{1}{1+a^2b^2}+\frac{1}{1+b^4}+\frac{1}{1+b^2c^2}+\frac{1}{1+c^4}\frac{1}{1+c^2a^2}+\frac{1}{1+a^4}\)
\(\Rightarrow2VT\ge\frac{2}{1+ab^3}+\frac{2}{1+bc^3}+\frac{2}{1+ca^3}\)
\(\Rightarrow VT\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a, b, c >0. Cm:
\(\frac{1}{a^4+1}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
BĐT sai khi \(a;b;c\) thuộc \(\left(0;1\right)\) và \(a;b;c\) không bằng nhau
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Chứng minh rằng với mọi a,b,c : \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\)≥\(\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
b) Chứng minh rằng:
\(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Bất đẳng thức \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}\ge\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\) của bạn sai với
[ a=1, b=1, c = 1/2 ]
Hãy check kỹ bất đẳng thức của bạn trước khi đăng câu hỏi nhé.
Cho a,b,c>=1.
CMR: \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}>=\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
cho a , b , c 0. Chứng minh các bất đẳng thức :
1, ab + bc + ca gesqrt{abc}left(sqrt{a}+sqrt{b}+sqrt{c}right)
2, frac{ab}{c}+frac{bc}{a}+frac{ac}{b}ge a+b+c
3, ab+frac{a}{b}+frac{b}{a}ge a+b+1
4, frac{a^3}{b}+frac{b^3}{c}+frac{c^3}{a}ge ab+bc+ca
5, frac{a}{bc}+frac{b}{ca}+frac{c}{ab}gefrac{1}{a}+frac{1}{b}+frac{1}{c}
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cho a , b , c >0. Chứng minh các bất đẳng thức :
1, ab + bc + ca \(\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2, \(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3, \(ab+\frac{a}{b}+\frac{b}{a}\ge a+b+1\)
4, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
5, \(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
1.
Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)
\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2.
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Cộng vế với vế:
\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3.
Từ câu b, thay \(c=1\) ta được:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
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4.
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
5.
\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)
Cộng vế với vế:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
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1. bđt được viết lại thành
\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
Theo bđt AM-GM thì :
\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)
Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)
Cộng vế với vế
=> \(2\left(ab+bc+ca\right)\ge2\left(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\right)\)
=> \(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)( đpcm )
Dấu "=" xảy ra <=> a=b=c
11 tháng 2 2020 lúc 21:04
1. left{{}begin{matrix}a,b,c0a+b+c1end{matrix}right.. Cmr: frac{ab}{sqrt{left(1-cright)^2left(1+cright)}}+frac{bc}{sqrt{left(1-aright)^2left(1+aright)}}+frac{ca}{sqrt{left(1-bright)^3left(1+bright)}}lefrac{3sqrt{2}}{8}
2. left{{}begin{matrix}a,b,c0a+b+cle1end{matrix}right.. Cmr: frac{1}{a^2+b^2+c^2}+frac{1}{ableft(a+bright)}+frac{1}{bcleft(b+cright)}+frac{1}{acleft(a+cright)}gefrac{87}{2}
3. left{{}begin{matrix}a,b,c0ab+bc+ca2abcend{matrix}right.. Cmr: frac{1}{aleft(2a-1right)^2}+frac{1}{bleft...
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1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!