cho a>b>c>0 và a^2+b^2+c^2=1. cmr a^3/(b+c)+b^3/(a+c)+c^3/(a+b)>=1/2
cho a b c 0 và a+b+c=3 CMR a/1+b^2 +b/1+c^2 +c/1+a^2 >=3/2
Ta có : \(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Đánh giá tương tự , ta cũng có :
\(\frac{b}{1+c^2}\ge b-\frac{bc}{2},\frac{c}{1+a^2}\ge c-\frac{ab}{2}\)
Từ đó suy ra :
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{ab+bc+c}{2}=3-\frac{ab+bc+ca}{2}\)
Mặt khác ,ta biết rằng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3.\)Từ đây ,kết hợp với đánh giá ở trên ,ta có kết quả cần chứng minh.
\(Ta\)\(có\) \(\frac{a}{1+b^2}\ge a-\frac{ab^2}{1+b^2}\)
Áp dụng bất đẳng thức \(a^2+b^2\ge2ab\)ta có
\(a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Chứng minh tương tụ với \(\frac{b}{1+c^2};\frac{c}{1+a^2}\)ta được
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge a+b+c-\frac{ab+bc+ac}{2}\) \(\left(1\right)\)
Mặt khác ta có :
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(Hay\)\(3^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow ab+bc+ca\le3\)\(\left(2\right)\)
\(Từ\)\(\left(1\right)\)\(\left(2\right)\)\(\Rightarrow\)\(a+b+c-\frac{ab+bc+ac}{2}\)\(\ge3-\frac{3}{2}=\frac{3}{2}\)\(\left(3\right)\)
\(Từ\)\(\left(1\right)\)\(\left(3\right)\)\(\Rightarrow\)\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\)
\(\left(đpcm\right)\)
Áp dụng BĐT AM-GM ta có:\(\frac{a}{1+b^2}=\frac{a\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)
Tương tự \(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
Kết hợp giả thiết và \(\left(ab+bc+ca\right)\le\frac{\left(a+b+c\right)^2}{3}\) ta có:
\(LHS\ge a+b+c-\frac{ab+bc+ca}{2}\ge a+b+c-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)
Đẳng thức xảy ra khi a=b=c=1
Cho a, b, c > 0 và a+b+c=3 . CMR :
\(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\)
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Cho: \(\left(a+b+c\right)^2=a^2+b^2+c^2\) và a, b, c khác 0. CMR: \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Cho: \(\left(a+b+c\right)^2=a^2+b^2+c^2\) và a,b, c khác 0. CMR: \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Từ đkđb
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Hớ hớ bài này mình cũng làm rồi.
Ta có: (a+b+c)2=a2+b2+c2
<=> a2+b2+c2+2(ab+bc+ca)=a2+b2+c2
<=>2(ab+bc+ca)=0
<=>ab+bc+ca=0
\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
=> \(\dfrac{1}{a^3}+\dfrac{3}{a^2b}+\dfrac{3}{ab^2}+\dfrac{1}{b^3}=-\dfrac{1}{c^3}\)
=>\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=\dfrac{3}{abc}\)
=> Đpcm.
cho (a+b+c)^2= a^2+b^2+c^2 và a,b,c # 0. CMR 1/a^2 + 1/b^2 + 1/c^2 = 3/abc
Bài5: cho a,b,c>0.CMR
1, 2/a+1/b >= 4/a+b
2, 1/a+1/b+1/c>= a/a+b+c
Bài 6: cho a,b>=0 cmr
1, a^3+b^4>=ab(a+b)
2, a^4+b^4>=ab(a^2+b^2)
3, a5+b5>=ab(a^3+b^3)
Bài 7 cho a,b,c>0 cmr
1/a^3+b^3+abc +1/b^3+c^3+abc+1/c^3+a^3+2 <1/abc
Bài 8cho a,b,c>0;abc=1
1, 1/a^3+b^3+2 +1/b^3+c^3+2 +1/c^3+a^3+2 =< 1
2,ab/a^5+b^5+ab +bc/b^5+c^5+bc + ca/c^5+a^5+ca =<1
Cho a,b,c khác 0 t/m (a+b+c)^2=a^2+b^2+c^2.CMR: 1/a^3+1/b^3+1/c^3=3/abc
ta có: (a+b+c)2 = a2 + b2 + c2
=> 2.(ab+ac+bc) = 0
ab + ac + bc = 0
=> 1/a + 1/b + 1/c = 0
Lại có: \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right).\)
\(=0.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)=0\)
=> 1/a3 + 1/b3 + 1/c3 -3/abc = 0
=> 1/a3 + 1/b3 + 1/c3 = 3/abc
a) Cho a2 + b2 + c2 + 3 = 2. (a + b + c)
CMR: a = b = c = 1
b) Cho (a + b + c)2 = 3. (ab + bc + ca)
CMR: a = b = c
c) Cho a + b + c = 0
CMR: a3 + b3 + c3 = 3abc
d) Cho a3 + b3 + c3 = 3abc
CMR: a + b + c = 0
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) (chuyển vế qua)
\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Do VP >=0 với mọi a, b, c. Nên để đăng thức xảy ra thì a = b = c
c) a + b + c = 0 suy ra a = -(b+c)
\(a^3+b^3+c^3=b^3+c^3-\left(b+c\right)^3\)
\(=b^3+c^3-b^3-3bc\left(b+c\right)-c^3\)
\(=3bc.\left[-\left(b+c\right)\right]=3abc\) (đpcm)
a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Do VT >=0 với mọi a, b, c nên a = b = c 1
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