Chứng minh:
a) (a+b+c)^2+(a+b-c)^2+(a-b+c)^2+(b+c-a)^2=4(a^2+b^2+c^2)
b) (ab+bc+ca)^2+(a^2-bc)+(b^2-ca)+(c^2-ab)=(a^2+b^2+c^2)^2
cho a,b,c là số thực dương. Cmr: a/b^2+ bc+c^2 + b/c^2+ ca+a^2 + c/ a^2+ ab+ b^2 >= a/ b^2+ bc + c^2 + b/c^2+ca+a^2 + c/a^2+ab + b^2 >= a+b+c/ab+ bc + ca.
\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)
chứng minh đẳng thức:
(a^2+b^2+c^2-ab-bc-ca).(a+b+c)=a(a^2-bc)+b(b^2-ca)+c(c^2-ab)
chứng minh : a^2+b^2+c^2-ab-b-ca)(a+b+c)= a(a^2-bc)+b(b^2-ca)+c(c^2-ab)
Ta có a(a2 - bc) + b(b2 - ca) + c(c2 - ab)
= a3 + b3 + c3 - 3abc
= (a + b)3 - 3ab(a + b) + c3 - 3abc
= [(a + b)3 + c3] - 3ab(a + b + c)
= (a + b + c)[(a + b)2 - (a + b)c + c2] - 3ab(a + b + c)
= (a + b + c)(a2 + b2 + c2 + 2ab - ac - bc - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - ac - bc) (đpcm)
cho 3 số thực dương a,b,c. chứng minh
\(ab+bc+ca\le\frac{a^3\left(b+c\right)}{a^2+bc}+\frac{b^3\left(c+a\right)}{b^2+ca}+\frac{c^3\left(a+b\right)}{c^2+ab}\le a^2+b^2+c^2\)\(ab+bc+ca\le\frac{a^3\left(b+c\right)}{a^2+bc}+\frac{b^3\left(c+a\right)}{b^2+ca}+\frac{c^3\left(a+b\right)}{c^2+ab}\le a^2+b^2+c^2\)
cho a,b,c>0, chứng minh:
1)ab+bc+ca >= a√ab+b√ca+c√ab
2)a^2+b^2+c^2 >= a√ab+b√ca+c√ab
1, Áp dụng BĐT cosi cho a,b,c>0
\(ab+bc\ge2\sqrt{ab^2c}=2b\sqrt{ac}\\ bc+ca\ge2\sqrt{abc^2}=2c\sqrt{ab}\\ ca+ab\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow2\left(ab+bc+ac\right)\ge2\left(b\sqrt{ac}+a\sqrt{bc}+c\sqrt{ab}\right)\\ \Leftrightarrow ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
\(2,\)
Ta có
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Áp dụng BĐT cm ở câu 1
Suy ra đpcm
Cho a,b,c>0. Cmr: a) \(\frac{ab}{a^2+bc+ca}+\frac{bc}{b^2+ca+ab}+\frac{ca}{c^2+ab+bc}\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)
b) \(\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le1\)
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
@tth_new, @Nguyễn Việt Lâm, @No choice teen, @Akai Haruma
giúp e vs ạ! Cần gấp
Thanks nhiều
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho a + b + c = 0. Chứng minh rằng:
a, ( ab + bc + ca ) 2 = a2b2 + b2c2 + c2a2
b, a ^ 4 + b ^ 4 + c ^ 4 = 2 x ( ab + bc + ca )2
Dùng hằng đang thuc la ra~~~daif qua nen ngai viet
p giúp mk câu b đk k? Mk đọc mãi cũng không hiểu lắm câu a thì làm đk r
$\rm Cho\ a,b,c \ge 0 .Thoả \ mãn \ ab+bc+ac=abc .Chứng \ minh\ a^{2}+b^{2}+c^{2}+5abc \ge 8$
`b)` Cho` a,b,c>=0,ab+bc+ca+abc=4`
CMR:`a^2+b^2+c^2+5abc>=8`
a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)
Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)
\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)
b.
\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)
\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)
\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)
- TH1: nếu \(a+b+c\ge4\)
Ta có: \(ab+bc+ca=4-abc\le4\)
\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)
(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)
- TH2: nếu \(3\le a+b+c< 4\)
Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)
\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)
Áp dụng BĐT Schur:
\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)
\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)
(Dấu "=" xảy ra khi \(a=b=c=1\))