R
rua_it
[tex]\mathrm{\sum_{cyclic}^{(a+c)(b+b)=1} \frac{1}{(a-b)^2} \geq ?[/tex]
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R
rua_it
[tex]\mathrm{\sum_{cyclic}^{a^2+b^2+c^2=3} \frac{1}{ab+1} \geq ?[/tex]
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B
bigbang195
Thui cho 1 bài BDT vào đây vậy
[TEX]a,b,c[/TEX] không âm. chứng minh
[TEX]a+b+c \ge 3\sqrt[3]{abc}[/TEX]
[TEX]a,b,c[/TEX] không âm. chứng minh
[TEX]a+b+c \ge 3\sqrt[3]{abc}[/TEX]
R
rua_it
[tex]a+b+c>0 \Rightarrow a+b+c+\frac{a+b+c}{3} \geq 2.\sqrt{ab}+2\sqrt{\frac{(a+b+c).c}{3}}[/tex]
[tex]\Rightarrow \frac{a+b+c}{3} \geq \frac{1}{2}[\sqrt{ab}+\sqrt{\frac{(a+b+c).c}{3}}] \geq 4.\sqrt{\sqrt{ab}.\sqrt{\frac{c.(a+b+c)}{3}}}[/tex]
[tex]\Rightarrow (\frac{a+b+c}{3})^4 \geq abc.[\frac{a+b+c}{3}] \Rightarrow \frac{a+b+c}{3} \geq \sqrt[3]{abc}[/tex]
[tex]\Rightarrow \frac{a+b+c}{3} \geq \frac{1}{2}[\sqrt{ab}+\sqrt{\frac{(a+b+c).c}{3}}] \geq 4.\sqrt{\sqrt{ab}.\sqrt{\frac{c.(a+b+c)}{3}}}[/tex]
[tex]\Rightarrow (\frac{a+b+c}{3})^4 \geq abc.[\frac{a+b+c}{3}] \Rightarrow \frac{a+b+c}{3} \geq \sqrt[3]{abc}[/tex]
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B
bigbang195
[tex]a+b+c>0 \Rightarrow a+b+c+\frac{a+b+c}{3} \geq 2.\sqrt{ab}+2\sqrt{\frac{(a+b+c).c}{3}}[/tex]
[tex]\Rightarrow \frac{a+b+c}{3} \geq \frac{1}{2}[\sqrt{ab}+\sqrt{\frac{(a+b+c).c}{3}}] \geq 4.\sqrt{\sqrt{ab}.\sqrt{\frac{c.(a+b+c)}{3}}}[/tex]
[tex]\Rightarrow (\frac{a+b+c}{3})^4 \geq abc.[\frac{a+b+c}{3}] \Rightarrow \frac{a+b+c}{3} \geq \sqrt[3]{abc}[/tex]
Chuẩn hóa hay hơn nhể.:-?
Không spam nữa àh bigbang.
cách này cũng được nhưng hơi dài :
[TEX]a+b+c+\sqrt[3]{xyz} \ge 4\sqrt[3]{xyz}[/TEX] OK !
tiếp
[TEX]a+b+c+d+e \ge 5\sqrt[5]{abcde}[/TEX]
khó chưa ! ^^)
Q
quyenuy0241
Chẳng khó gì :
3 số:
[tex]a+b+c \ge 3\sqrt[3]{abc}[/tex]
2 số[tex]d+e\ge 2\sqrt{de}[/tex]
[tex]a+b+c+d+e \ge \sqrt[3]{abc}+\sqrt[3]{abc}+\sqrt[3]{abc}+\sqrt{de}+\sqrt{de} \ge 5 \sqrt[5]{abcde}[/tex]
B
bigbang195
Q
quyenuy0241
Sorry em nha :Bảo chứng minh AM-GM 5 biến lại đi xài luôn /
nhầm:làm lại nhé:
6 số thì OK roài nhá:
[tex]a+b+c+d+e+\frac{a+b+c+d+e}{5}\geq6\sqrt[6]{\frac{abcde(a+b+c+d+e)}{5}}[/tex]
Hay[tex] a+b+c+d+e\geq5\sqrt[5]{abcde}[/tex]
đây chính là cơ sở để CM AM-GM nhiều số:
R
rua_it
[tex](gt) \Rightarrow a^2+b^2+c^2+abc=4 \Rightarrow (a,b,c) \in\ (0;2)[/tex]
[tex]\Rightarrow27.(2-a)(2-b)(2-c) \leq_{AM-GM} (2-a+2-b+2-c)^3[/tex]
[tex]\Rightarrow27.[8+2(ab+bc+ca)+a^2+b^2+c^2-4-4(a+b+c)] \leq [6-(a+b+c)]^3[/tex]
[tex]\Rightarrow27.[(a+b+c)^2-4.(a+b+c)+4] \leq [6-(a+b+c]^3[/tex]
[tex]\Rightarrow[(a+b+c)^2+12.(a+b+c)+36].[(a+b+c)-3] \leq 0[/tex]
[tex] f(x)=[(a+b+c)^2+12.(a+b+c)+36] > 0, \forall a+b+c[/tex]
[tex]\Rightarrow a+b+c-3 \leq0 [/tex]
[tex]\Rightarrow a+b+c \leq3(dpcm)[/tex]
R
rua_it
[tex](gt) \rightarrow (a,b,c) \in\ (0;2)[/tex] [tex] A;B \in\ (0;\frac{\pi}{2})[/tex]
[tex] \rightarrow \left{\begin{a=2cosA}\\{b=2cosB}[/tex]
[tex]a^2+b^2+c^2+abc=4 \rightarrow c=\frac{\sqrt{(4-a^2).(4-b^2)}-ab}{2}[/tex]
[tex]=\frac{4sinAsinB-4cosAcosB}{2}=2cos[\pi-(A+B)]=2cosC[/tex]
[tex] a+b+c \leq 3 \rightarrow 2cosA+2cosB+2cosC \leq 3 [/tex]
[tex]\rightarrow \sum_{cyc} cosA \leq \frac{3}{2}[/tex]
[tex]\rightarrow 2cos.\frac{A+B}{2}cos.\frac{A-B}{2} -2cos^2.\frac{A+B}{2}+1 \leq \frac{3}{2}[/tex]
[tex]\rightarrow cos^2.\frac{A+B}{2}+\frac{1}{4}-cos.\frac{A+B}{2}cos.\frac{A-B}{2} \geq 0[/tex]
[tex]\rightarrow (cos.\frac{A+B}{2}-\frac{1}{2}.cos.\frac{A-B}{2})^2+\frac{1}{4}.sin^2.\frac{A-B}{2} \geq 0[/tex] :đúng.
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S
sieuthamtu_sieudaochit
Lâu rồi không trở lại diễn đàn
[TEX]LHS-RHS=\frac{1}{2}( \sqrt[3]{a}+ \sqrt[3]{b}+ \sqrt[3]{c})\sum (\sqrt[3]{a}-\sqrt[3]{b})^2 \ge 0[/TEX]
L
lamtrang0708
bài này còn có thê lâp phương lên sau đó trư` vê' phân tích thành (a+b+c)(a2+b2+c2 - ab -bc - ca) luôn \geq0 do a,b,c ko âm và a2 + b2+ c2 \geq ab + bc+ ca