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giải đáp
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Nghiên cứu cái này khó quá :v
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giả sử $a\geq b\geq c$ khi đó $b^{2}-bc+c^{2}=b^{2} +c(c-b)\leq b^{2}$ $c^{2}-ac+c^{2}=c(c-a)+a^{2} \leq a^{2}$ $\Rightarrow P\leq a^{2}b^{2}(a^{2}-ab+b^{2})=\frac{4}{9}.\frac{3ab}{2}.\frac{3ab}{2}(a^{2}-ab+b^{2})$ $\leq \frac{4}{9}( \frac{\frac{3ab}{2}+\frac{3ab}{2}+(a^{2}-ab+b^{2})}{3})^{3}=\frac{4}{9.27}(a+b)^{6}\leq \frac{4}{9.27}(a+b+c)^{6}=12$ Dấu "=" $ a=2;b=1;c=0$ |
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sửa đổi
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(7)
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ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c= \min\left{ a{,}b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$≥4(a−c)(b−c)≥4ab+bc+ca
ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c$ là số nhỏ nhất, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$
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sửa đổi
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(7)
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ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c=min\left{ a,b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$
ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c= \min\left{ a{,}b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$≥4(a−c)(b−c)≥4ab+bc+ca
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sửa đổi
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(7)
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Ta có (a−b)(b−c)(c−a)≠0" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">(a−b)(b−c)(c−a)≠0(a−b)(b−c)(c−a)≠0 nên 0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)0=(a−b)+(b−c)+(c−a)=(a−b)+(b−c)+(c−a)(a−b)(b−c)(c−a)=∑1(a−b)(b−c)Do đó H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 33.898em; min-height: 0px; padding-top: 1px; padding-bottom: 1px; width: 745px; position: relative;">H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)H=∑1(a−b)2=∑1(a−b)2+2∑1(a−b)(b−c)=(∑1a−b)2=(1a−b+a−b(a−c)(b−c))2=1(a−b)2+(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)Không mất tính tổng quát, giả sử c=min{a,b,c}" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">c=min{a,b,c}c=min{a,b,c}, khi đó ab+bc+ca⩾(a−c)(b−c)⩾0" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">ab+bc+ca⩾(a−c)(b−c)⩾0ab+bc+ca⩾(a−c)(b−c)⩾0 nên:H⩾21(a−b)2.(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)⩾4(a−c)(b−c)⩾4ab+bc+ca=1" role="presentation" style="display: inline-block; line-height: 0; font-size: 18.06px; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">H⩾2√1(a−b)2.(a−b)2(a−c)2(b−c)2+2(a−c)(b−c)⩾4(a−c)(b−c)⩾4ab+bc+ca
ta có $(a-b)(b-c)(c-a)\neq0$ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$=$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$giả sử $c=min\left{ a,b,c \right}$, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi)đấu "=" $\Leftrightarrow ......$
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giải đáp
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(7)
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ta có $(a-b)(b-c)(c-a)\neq0$ ta có $0=a-b+b-c+c-a=\frac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}=\sum\frac{1}{(a-b)(b-c)}$ H=$\sum\frac{1}{(a-b)^{2}}=\sum\frac{1}{(a-b)^{2}}+2\sum_{a}^{b} \frac{1}{(a-b)(b-c)}$ =$ (\sum\frac{1}{a-b})^{2}=\frac{1}{(a-b)^{2}}+(\frac{(a-b)^{2}}{((b-c)(a-c))^{2}}+2\frac{a-b}{(a-c)(b-c)}$ giả sử $c$ là số nhỏ nhất, khi đó $ab+bc+ca \geq (a-c)(b-c)\geq0$ H$\geq \frac{4}{(a-c)(b-c)}\geq \frac{4}{ab+bc+ca}$ ( BĐT cosi) đấu "=" $\Leftrightarrow ......$
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giải đáp
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(5)
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ad BĐT C-S ta có $(a^{2}+b+c)(1+b+c)\geq (a+b+c)^{2}$ $\Rightarrow \sqrt{\frac{a^{2}}{a^{2}+b+c}}\leq \frac{a\sqrt{b+c+1}}{a+b+c}=\frac{a\sqrt{3(b+c+1)}}{(a+b+c)\sqrt{3}}\leq \frac{a+(b+c+1+3)}{2(a+b+c)\sqrt{3}}$ TT $\Rightarrow VT\leq \frac{4(a+b+c)+2(ab+bc+ca)}{2(a+b+c)\sqrt{3}}=\frac{2\sqrt{3}}{3}+\frac{1}{\sqrt{3}} \frac{ab+bc+ca}{a+b+c}=\frac{2\sqrt{3}}{3}+\frac{1}{3}\sqrt{ab+bc+ca}\leq \sqrt{3}$ Dấu "=" $\Leftrightarrow a=b=c=1$ |
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được thưởng
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Đăng nhập hàng ngày 24/08/2016
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giải đáp
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Bất đẳng thức 3(ACAMOPHOMADADY 2016-2017)
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2. ta có $\frac{x}{x^{4}+1+2xy} \leq \frac{x}{2x^{2}+2xy}=\frac{1}{2}.\frac{1}{x+y}\leq \frac{1}{8}(\frac{1}{x}+\frac{1}{y})$ TT $\Rightarrow VT \leq \frac{1}{4}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{4}$ Dấu "=" $\Leftrightarrow x=y=z=1$ |
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