B
bigbang195
Last edited by a moderator:
- Status
C
chinhphuc_math
B
bigbang195
V
vodichhocmai
[TEX](b+c)^2-a(b+c)+\frac{a^3}{3}-3bc>0[/TEX]
[TEX]\Delta_{b+c}=\frac{36-a^3}{3a}<0[/TEX]
[TEX]Done!![/TEX]
N
namtuocvva18
Cho a,b,c dương và [TEX]a+b+c=6[/TEX]. Chứng minh:
[TEX]\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+ \frac{c}{\sqrt{a^3+1}}\geq 2[/TEX].
[TEX]\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+ \frac{c}{\sqrt{a^3+1}}\geq 2[/TEX].
N
namtuocvva18
Cho a,b,c là độ dai 3 cạnh tam giác. Chung minh:
[TEX](3-\frac{b+c}{a})(3-\frac{c+a}{b})(3-\frac{a+b}{c})\leq 1[/TEX].
[TEX](3-\frac{b+c}{a})(3-\frac{c+a}{b})(3-\frac{a+b}{c})\leq 1[/TEX].
N
namtuocvva18
Cho x,y,z dương. Tìm GTNN của:
[TEX]P=\frac{x^4}{y^4}+\frac{y^4}{x^4}-\frac{x^2}{y^2}-\frac{y^2}{x^2}+\frac{x}{y}+\frac{y}{x}[/TEX].
[TEX]P=\frac{x^4}{y^4}+\frac{y^4}{x^4}-\frac{x^2}{y^2}-\frac{y^2}{x^2}+\frac{x}{y}+\frac{y}{x}[/TEX].
N
namtuocvva18
Cho a,b,c dương và [TEX]3a^2+2b^2+c^2=1[/TEX]. Tìm GTNN của:
[TEX]P=\frac{3a}{bc}+\frac{4b}{ca}+\frac{5c}{ab}[/TEX].
[TEX]P=\frac{3a}{bc}+\frac{4b}{ca}+\frac{5c}{ab}[/TEX].
N
namtuocvva18
Cho a,b,c dương và [TEX]a+b+c=1[/TEX]. Chứng minh:
[TEX]\sqrt{\frac{ab}{c}+1}+\sqrt{\frac{bc}{a}+1}+\sqrt{\frac{ca}{b}+1}\geq 2(\sqrt{a}+\sqrt{b}+\sqrt{c})[/TEX].
[TEX]\sqrt{\frac{ab}{c}+1}+\sqrt{\frac{bc}{a}+1}+\sqrt{\frac{ca}{b}+1}\geq 2(\sqrt{a}+\sqrt{b}+\sqrt{c})[/TEX].
N
namtuocvva18
Môt cach giai
Cho a,b,c dương. Chứng minh:
[TEX]\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}.[/TEX]
Cho a,b,c dương. Chứng minh:
[TEX]\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}.[/TEX]
N
namtuocvva18
Giai:
Ta có:
[TEX]\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}=\frac{a^2-ab+b^2}{b}+\frac{b^2-bc+c^2}{c}+\frac{c^2-ca+a^2}{a}=A[/TEX]
Lai co:
[TEX](a+b+c).A\geq (\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2})^2[/TEX]
Và
[TEX]\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\geq a+b+c[/TEX]
Suy ra:
[TEX] (a+b+c).(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq (a+b+c)(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2})[/TEX]
=>dpcm.
Last edited by a moderator:
V
vodichhocmai
[TEX]LHS:=\sum_{cyc}\sqrt{\frac{ab}{c}+a+b+c} =\sum_{cyc}\sqrt{\frac{(c+a)(c+b)}{c}[/TEX]
[TEX](c+a)(b+c)\ge c\(\sqrt{a}+\sqrt{b}\)^2[/TEX]
[TEX]\righ LHS\ge 2 (\sqrt{a}+\sqrt{b}+\sqrt{c}) [/TEX]
N
namtuocvva18
Cho a,b,c dương và [TEX]ab+bc+ca=3[/TEX]. Chứng minh:
[TEX]\frac{1}{1+a^2(b+c)}+\frac{1}{1+b^2(c+a)}+\frac{1}{1+c^2(a+b)}\leq\frac{1}{abc}[/TEX].
[TEX]\frac{1}{1+a^2(b+c)}+\frac{1}{1+b^2(c+a)}+\frac{1}{1+c^2(a+b)}\leq\frac{1}{abc}[/TEX].
N
namtuocvva18
N
namtuocvva18
Cho [TEX]x,y,z [/TEX][TEX]\geq 0[/TEX] [TEX]x+y+x=3[/TEX]
Cm:[TEX]\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+ \frac{1}{\sqrt{z}} \geq 3[/TEX]
Cm:[TEX]\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+ \frac{1}{\sqrt{z}} \geq 3[/TEX]
B
bigbang195
[TEX]VT \ge \frac{9}{\sum \sqrt{x}} \ge 3[/TEX] .
Last edited by a moderator:
B
bigbang195
[TEX]x^3+y^3+1=\frac{1}{2}(x+y+1)((x-1)^2+(y-1)^2+(x-y)^2) +3xy \ge 3xy[/TEX]
B
bigbang195
[TEX]ab+bc+ac \ge 3\sqrt[3]{abc^2} \Rightarrow abc \le 1[/TEX]
nên
[TEX]VT \le \frac{1}{abc+a^2(b+c)}+\frac{1}{abc+b^2(a+c)}+\frac{1}{abc+c^2(a+b}) =\frac{1}{a(ab+bc+ac)}+\frac{1}{b(ab+bc+ac)}+\frac{1}{c(ab+bc+ac)}=\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{3}.\frac{ab+bc+ac}{abc}=\frac{1}{abc}=VP[/tex]
Last edited by a moderator:
Q
quyenuy0241
Sax nhầm đề roài thế kia thì nói làm quái gì phải thế này này:
[tex](1+\frac{b-c}{a})^a.(1+\frac{c-a}{b})^b.(1+\frac{a-b}{c})^c\le 1[/tex]
Q
quyenuy0241
[TEX]\frac{a}{\sqrt{b^3+1}}=\frac{a}{\sqrt{(b+1)(b^2-b+1)}}\ge \frac{2a}{b^2+2}[/TEX]
Các bất đẳng thức sau tương tự
Vậy cần CM [tex]\sum{\frac{2a}{{b^2+2}}\ge 2[/tex] Tới đây có thể dùng AM-GM ngược dấu đảm bảo ra!!!/
//- Status