CMR:
\(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\)
\(\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\)
\(\)
Cho a, b, c > 0. CMR:
a, \(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge2\left(a+b+c\right)\)
b, \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
c, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
Giúp mình với các bạn ơiii
a) Bổ đề: \(x^3+y^3\ge xy\left(x+y\right)\forall x,y>0\)
\(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge\frac{ab\left(a+b\right)}{ab}+\frac{bc\left(b+c\right)}{bc}+\frac{ca\left(c+a\right)}{ca}=2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Cảm ơn bạn nhiều nhé Nhật Pháp soi chiếu thế gian. Nếu có thể, mong bạn hãy giúp mình những phần còn lại ^^
c) Áp dụng bất đẳng thức AM-GM:
\(\frac{a^3}{b}+\frac{a^3}{b}+b^2\ge3\sqrt[3]{\frac{a^3}{b}.\frac{a^3}{b}.b^2}=3a^2\);
\(\frac{b^3}{c}+\frac{b^3}{c}+c^2\ge3\sqrt[3]{\frac{b^3}{c}.\frac{b^3}{c}.c^2}=3b^2\);
\(\frac{c^3}{a}+\frac{c^3}{a}+a^2\ge3\sqrt[3]{\frac{c^3}{a}.\frac{c^3}{a}.a^2}=3c^2\)
Cộng theo từng vế ba bất đẳng thức trên ta đươc:
\(2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+a^2+b^2+c^2\ge3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
Dấu "=" xảy ra khi \(a=b=c\)
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.
áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
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:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
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:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
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
cho a, b, c>0. CMR a\(\frac{a^3}{b}\ge a^2+ab-b^2\)
CM \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)
Cho a, b, c là độ dài 3 cạnh của tam giác CM \(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Tự nhiên lục được cái này :'(
3. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{\left(1+1\right)^2}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\)
\(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{b+c-a+c+a-b}=\frac{4}{2c}=\frac{2}{c}\)
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{\left(1+1\right)^2}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)
Cộng theo vế ta có điều phải chứng minh
Đẳng thức xảy ra <=> a = b = c
câu 1 :Cmr a)\(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)
b) \(\frac{a^3+b^3+c^3}{3}\ge\left(\frac{a+b+c}{3}\right)^3\)
câu 2 : cho a+b=1 .Cm \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
câu 3: cho a+b+c=1và a,b,c>0.CMR \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
câu 4 Tim max của : ab+2(a+b) ...biết a2+b2=1
Câu 2)
Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)
\(\Leftrightarrow9\ge4\left(ab+2\right)\)
\(\Rightarrow9\ge4ab+8\)
\(\Rightarrow1\ge4ab\)
Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow a^2+2ab+b^2\ge4ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )
Câu 3)
Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
Mà \(a+b+c=1\)
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)
\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si
\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)
\(\Rightarrow\) ĐPCM
\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
Cho a,b,c>0 CMR
\( \frac{a^3}{bc}+ \frac{b^3}{ac}+ \frac{c^3}{ab}\ge \frac{3(a^2+b^2+c^2)}{a+b+c} \)
Cho a,b,c > 0. CMR:
1. \(a^3+b^3+c^3\ge3abc\)
2. \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)
3. \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)
4. \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
5. \(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}\ge\frac{1}{ab+1}\)
6.\(\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3}\ge\frac{3}{1+abc}\)