Cho a,b,c>0,ab+bc+ca=abc
CMR:1/a+1/b+1/c+5(a+b+c)<3/16
cho a,b,c >0 thỏa mãn ab + bc + ca = abc
CMR: \(\dfrac{\sqrt{b^2+2a^2}}{ab}+\dfrac{\sqrt{c^2+2b^2}}{bc}+\dfrac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\)
Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).
Đặt biểu thức ở VT là A. Ta có:
\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).
Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).
Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.
Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).
Đẳng thức xảy ra khi a = b = c = 3.
Vậy...
Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)
\(VT\ge\dfrac{\sqrt{\dfrac{1}{3}\left(a+a+b\right)^2}}{ab}+\dfrac{\sqrt{\dfrac{1}{3}\left(b+b+c\right)^2}}{bc}+\dfrac{\sqrt{\dfrac{1}{3}\left(c+c+a\right)^2}}{ca}\)
\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=3\)
cho a, b,c >0 thỏa mãn ab+bc+ca=abc
CMR : (√b2+2a2)/ab + (√c2+2b2)/bc + (√a2+2c2)/ac
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 thoả mãn a+b+c=1
chứng minh rằng √(a+bc) +√(b+ca) +√(c+ab)≥1+√bc+√ca+√ab
cho a,b,c>0 thoả mãn a+b+c=1
chứng minh rằng √(a+bc) +√(b+ca) +√(c+ab)≥1+√bc+√ca+√ab
Ta chứng minh:\(\sqrt{a+bc}\ge a+\sqrt{bc}\)
\(\Leftrightarrow a+bc\ge a^2+bc+2a\sqrt{bc}\)
\(\Leftrightarrow a\ge a^2+2a\sqrt{bc}\)\(\Leftrightarrow a\ge a\left(a+2\sqrt{bc}\right)\Leftrightarrow1\ge a+2\sqrt{bc}\Leftrightarrow a+b+c\ge a+2\sqrt{bc}\)
\(\Leftrightarrow b+c-2\sqrt{bc}\ge0\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)(luôn đúng)
\(\Leftrightarrow\sqrt{a+bc}\ge a+\sqrt{bc}\)
CMTT\(\sqrt{b+ca}\ge b+\sqrt{ca}\)
\(\sqrt{c+ab}\ge c+\sqrt{ab}\)
\(\Leftrightarrow\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)Vậy ......
(Dấu = xảy ra (=) a=b=c=1/3
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>0 và 1/ab+1/bc+1/ca>=1. cmr: a/bc+b/ca+c/ab>=1
Cho a,b,c >0 tm abc=1, C/m
\(\dfrac{1}{\sqrt{a^5+b^2+ab+6}}+\dfrac{1}{\sqrt{b^5+c^2+bc+6}}+\dfrac{1}{\sqrt{c^5+a^2+ca+6}}\le1\)
\(a^5+b^2+ab+6\ge3a^2b+6\)
\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)
\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)
\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)
\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)
\(\Rightarrow P\le\sqrt{1}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
cho a,b,c>=0, a+b+c=1. chứng minh rằng (a-bc)/(a+bc)+(b-ca)/(b+ca)+(c-ab)/(c+ab)<=3/2
\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)
\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)
Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)
=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)
\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)