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1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
B1:
\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Xét hiệu:
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
=> BĐT luôn đúng
*
Ta có:
\(a< b+c\Rightarrow a^2< ab+ac\)
\(b< a+c\Rightarrow b^2< ab+ac\)
\(c< a+b\Rightarrow a^2< ac+bc\)
Cộng từng vế bất đẳng thức ta được:
\(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Vậy: \(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
B2:
Ta có: \(a+b>c\) ; \(b+c>a\); \(a+c>b\)
Xét:\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{a+b+c}+\dfrac{1}{a+c+b}=\dfrac{2}{a+b+c}>\dfrac{2}{b+c+b+c}=\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+c+a+c}=\dfrac{1}{a+c}\)
Suy ra:
\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
=> ĐPCM
1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi
1. Ta có: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Đặt \(\hept{\begin{cases}\frac{1}{a}=m\\\frac{1}{b}=n\\\frac{1}{c}=p\end{cases}}\) khi đó \(\hept{\begin{cases}m+n+p=3\\M=2\left(m^2+n^2+p^2\right)+mnp\end{cases}}\)
Áp dụng Cauchy ta được:
\(\left(m+n-p\right)\left(m-n+p\right)\le\left(\frac{m+n-p+m-n+p}{2}\right)^2=m^2\)
\(\left(n+p-m\right)\left(n+m-p\right)\le n^2\)
\(\left(p-n+m\right)\left(p-m+n\right)\le p^2\)
\(\Rightarrow\left(m+n-p\right)\left(n+p-m\right)\left(p+m-n\right)\le mnp\)
\(\Leftrightarrow m^3+n^3+p^3+3mnp\ge m^2n+mn^2+n^2p+np^2+p^2m+pm^2\)
\(\Leftrightarrow\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-pm\right)+6mnp\ge mn\left(m-n\right)+np\left(n-p\right)+pm\left(p-m\right)\)
\(=mn\left(3-p\right)+np\left(3-m\right)+pm\left(3-n\right)\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)-3\left(mn+np+pm\right)+6mnp\ge3\left(mn+np+pm\right)-3mnp\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)+9mnp\ge6\left(mn+np+pm\right)\)
\(\Leftrightarrow xyz\ge\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(\Rightarrow M\ge2\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(=\frac{5}{3}\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m^2+n^2+p^2+2mn+2np+2pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m+n+p\right)^2\)
\(\ge\frac{4}{3}\cdot3+\frac{1}{3}\cdot3^2=4+3=7\)
Dấu "=" xảy ra khi: \(m=n=p=1\Leftrightarrow a=b=c=1\)
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
b,c tương tự
d)Áp dụng bđt AM-GM ta được
\(a^4+a^4+b^4+c^4\ge4\sqrt[4]{a^4a^4b^4c^4}=4a^2bc\)
TT\(\Rightarrow a^4+b^4+b^4+c^4\ge4ab^2c\)
\(a^4+b^4+c^4+c^4\ge4abc^2\)
Cộng vế theo vế ta được \(4\left(a^4+b^4+c^4\right)\ge4\left(a^2bc+ab^2c+abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\left(đpcm\right)\)
d)
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-ab^2c-abc^2\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2+2a^2c^2-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)+\left(b^2c^2+c^2a^2-2c^2abc\right)+\left(a^2b^2+c^2a^2-2a^2ab\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ac\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a , b , c
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
câu a (a+b+c)2 +(a+b-c)2 - 4c2= (a+b+c)2+(a+b-c+2c).(a+b-c-2c) =(a+b+c)2 +(a+b+c).(a+b-3c)=(a+b+c). (a+b+c+a+b-3c)=(a+b+c).2.(a+b-c)
câu b 4a2b2-(a2+b2-c2) = (2ab-a2-b2+c2).(2ab+a2+b2-c2)
= (c2-(a-b)2).((a+b)2-c2)
= (c-a+b).(c+a-b).(a+b-c).(a+b+c)
câu c a4+b4+c4-2a2b2+2b2c2-2a2c2-4b2c2=(a2-b2-c2)2-4b2c2=(a2-b2-c2-2bc).(a2-b2-c2+2bc)=(a2-(b+c)2).(a2-(b-c)2)=(a-b-c).(a+b+c).(a-b+c).(a+b-c)
câu d dùng pp xét giá trị riêng thay b =c (bạn tự giải ) thì đa thức này nếu coi là đa thức biến b thì đa thức A chia hết cho b-c
a,b,c bình đẳng => A chia hết cho c-a , a-b
=>A= k(a-b)(b-c)(c-a)
thay thử một bộ a,b,c bất kì => k=? (mình đang vội )
thay k tính đc vàoA= k(a-b)(b-c)(c-a)