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อ้างอิง:
72. Iran 1998 $z,y,z>1,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ $$\sqrt{x+y+z}\geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$ This one is tricky!! 74. APMO 1998 (a,b,c>0)Observe that $$K = \frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1.$$ Then Cauchy-Schwarz inequality implies $\displaystyle{ \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}=\sqrt{\frac{x-1}{x}}\cdot \sqrt{x}+\sqrt{\frac{y-1}{y}}\cdot \sqrt{y}+\sqrt{\frac{z-1}{z}}\cdot\sqrt{z}}$ $\leq \sqrt{K(x+y+z)}=\sqrt{x+y+z}.$ $$\Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big)\geq 2(1+\frac{a+b+c}{\sqrt[3]{abc}})$$ Let $\displaystyle{x=\frac{a}{\sqrt[3]{abc}},y=\frac{b}{\sqrt[3]{abc}},z=\frac{c}{\sqrt[3]{abc}}}$. Then we have $xyz=1$ and the inequality is equivalent to $$\Big(1+\frac{x}{y}\Big)\Big(1+\frac{y}{z}\Big)\Big(1+\frac{z}{x}\Big)\geq 2(1+x+y+z)$$ which is also equivalent to $$x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\geq 2(x+y+z).$$ By AM-GM inequality we have $x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\geq 6xyz.$ Thus $3(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2)\geq 2(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2 + 3xyz)$ $=2(x+y+z)(xy+yz+zx)$ $\geq 6(x+y+z)$ (because $xy+yz+zx\geq 3$ by AM-GM inequality). Therefore, $x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\geq 2(x+y+z)$, as required. |
76. Korea 1998 $(x,y,z>0,x+y+z=xyz)$
$$\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\geq \frac{3}{2}$$ Let $x=\tan{A},y=\tan{B},z=\tan{C}$. Then the inequality is equivalent to $$\cos{A}+\cos{B}+\cos{C}\leq \frac{3}{2}.$$ Note that we have $\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C} \Leftrightarrow A+B+C=\pi$. Since $\cos{x}$ is concave on $[0,\frac{\pi}{2}]$, Jensen's inequality implies that $$\cos{A}+\cos{B}+\cos{C}\leq 3\cos{\Big(\frac{A+B+C}{3}\Big)}=3\cos{\frac{\pi}{3}}=\frac{3}{2}.$$ |
78. IMO Shortlist 1998 $(x,y,z>0,xyz=1)$
$$\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+z)(1+x)}+\frac{z^3}{(1+x)(1+y)}\geq \frac{3}{4}$$ First, note that $x+y+z\geq 3$ by AM-GM inequality and $(x^{3/2}+y^{3/2}+z^{3/2})^2\geq \frac{1}{3}(x+y+z)^3$ by power mean inequality with parameters $\frac{3}{2},1$. Then we have by Cauchy-Schwarz inequality 80. Ireland 1997 $(a,b,c\geq 0,a+b+c\geq abc)$$\displaystyle{ LHS \geq \frac{(x^{3/2}+y^{3/2}+z^{3/2})^2}{3+2(x+y+z)+(xy+yz+zx)}}$ $\displaystyle{ \geq \frac{\frac{1}{3}(x+y+z)^3}{3(x+y+z)+\frac{1}{3}(x+y+z)^2}}$ $\displaystyle{ =\frac{x+y+z}{4}}$ $\displaystyle{ \geq \frac{3}{4}}$. $$a^2+b^2+c^2\geq abc.$$ Note that $abc\leq a+b+c$ implies $$3a^2b^2c^2 \leq 3abc(a+b+c)\leq (ab+bc+ca)^2 \leq (a^2+b^2+c^2)^2.$$ Thus $a^2+b^2+c^2 \geq \sqrt{3}abc \geq abc.$ |
81. Iran 1997 $a,b,c,d>0, abcd=1$
$$a^3+b^3+c^3+d^3 \geq \max\{a+b+c+d,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\}$$ By AM-GM inequality we have $$a^3+b^3+c^3\geq 3abc$$ $$b^3+c^3+d^3\geq 3bcd$$ $$c^3+d^3+a^3\geq 3cda$$ $$d^3+a^3+b^3\geq 3dab$$ Adding them together we get $$a^3+b^3+c^3+d^3 \geq abc+bcd+cda+dab = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$$ By AM-GM inequality we have $$a+b+c+d\geq 4$$ By power mean inequality we get $$a^3+b^3+c^3+d^3 \geq \frac{1}{16}(a+b+c+d)^3\geq \frac{(a+b+c+d)^3}{(a+b+c+d)^2}=a+b+c+d.$$ Thus we have the inequality as wanted. |
สู้ๆ คับเกือบเสร็จแล้วอีกหน่อยเดียวว เหะๆ ^^
ว่าแต่ทำไงให้เก่ง หนออออ |
อ้างอิง:
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82. Hong Kong 1997 $(x,y,z>0)$
$$\frac{3+\sqrt{3}}{9}\geq \frac{xyz(x+y+z+\sqrt{x^2+y^2+z^2})}{(x^2+y^2+z^2)(xy+yz+zx)}$$ We will prove two inequalities separately. 83. Belarus 1997 $(a,b,c>0)$$(1). \,\, 3xyz(x+y+z)\leq (xy+yz+zx)^2 \leq (xy+yz+zx)(x^2+y^2+z^2)$ $(2). \,\, 3\sqrt[3]{xyz}\leq x+y+z \leq \sqrt{3(x^2+y^2+z^2)}$ $3\sqrt[3]{x^2y^2z^2}\leq xy+yz+zx$ Thus $9xyz\leq (xy+yz+zx)\Big(\sqrt{3(x^2+y^2+z^2)}\Big)$ Using these two inequalities we get the result. $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{c+a}{b+c}$$ [This proof is incomplete] WLOG,assume $a\geq b\geq c.$ Then $\Big(\dfrac{a}{b}-\dfrac{c+a}{b+c}\Big)+\Big(\dfrac{b}{c}-\dfrac{a+b}{c+a}\Big)+\Big(\dfrac{c}{a}-\dfrac{b+c}{a+b}\Big)$ $= \dfrac{c(a-b)}{b(b+c)}+\dfrac{a(b-c)}{c(c+a)}+\dfrac{b(c-a)}{a(a+b)}$ $\geq \dfrac{c(a-b)}{a(a+b)}+\dfrac{a(b-c)}{a(a+b)}+\dfrac{b(c-a)}{a(a+b)}$ $=0.$ New Proof: The inequality is equivalent to $\Big(\dfrac{a}{b}-\dfrac{c+a}{b+c}\Big)+\Big(\dfrac{b}{c}-\dfrac{a+b}{c+a}\Big)+\Big(\dfrac{c}{a}-\dfrac{b+c}{a+b}\Big)\geq 0$ $\dfrac{c(a-b)}{b(b+c)}+\dfrac{a(b-c)}{c(c+a)}+\dfrac{b(c-a)}{a(a+b)}+3\geq 3$ $\dfrac{c^2+ab}{c^2+ca}+\dfrac{a^2+bc}{a^2+ab}+\dfrac{b^2+ca}{b^2+bc}\geq 3$ which is exactly #64. |
คับ จะพยายฝึกฝนฝีมือ ตอนนี้ก็ซื้อ อสมการ และ สมการเชิงฟังก์ชันมาอ่าน :sung: :sung: แต่พอเห็นเนื้อหา :died: :died: :aah: :aah: :cry: :cry: แทบช็อค!!!!
อยากให้มี ร่วมกันเฉลย หนังสือเล่นนี้จัง เหะๆ :happy: เพราะทำไม่ค่อยจะได้ :sweat: :sweat: ไว้มีอะไรไม่ได้เด่ว โพส ถามละกันดีกว่า :D :D :D |
หนังสือของโครงการสอวน. เหมาะสำหรับนักเรียนที่สอบผ่านสอวน. รอบแรกไปแล้วครับ จึงไม่แปลกอะไรที่คนทั่วไปจะเห็นว่ายาก โจทย์อสมการในหนังสือเล่มนี้เป็นโจทย์ระดับข้อสอบแข่งขันระหว่างประเทศ ข้อสอบคัดเลือกตัวแทนโอลิมปิกของแต่ละประเทศ รวมไปถึงโจทย์ที่รวบรวมจากวารสารคณิตศาสตร์ต่างๆ ระดับความยากจึงไม่ต้องพูดถึงครับ ใจเย็นๆครับ ค่อยๆอ่านไป ถ้าไม่เข้าใจตรงส่วนไหนก็เข้ามาถามได้ครับ :)
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84. The solution is here http://www.mathcenter.net/forum/show...?t=1186&page=3
85. Romania 1997 $(xyz=1,x,y,z>0)$ $$\frac{x^9+y^9}{x^6+x^3y^3+y^6}+\frac{y^9+z^9}{y^6+y^3z^3+z^6}+\frac{z^9+x^9}{z^6+z^3x^3+x^6}\geq 2$$ Let $a=x^3,b=y^3,c=z^3.$ Then $abc=1$ and the inequality becomes $$\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+bc+c^2}+\frac{c^3+a^3}{c^2+ca+a^2}\geq 2$$ Now we have $a^2+ab+b^2 \leq a^2+b^2+\dfrac{a^2+b^2}{2}=\dfrac{3}{2}(a^2+b^2)$ and $\dfrac{a^3+b^3}{a^2+b^2}\geq \sqrt{ab}$. Thus $$\frac{a^3+b^3}{a^2+ab+b^2}\geq \frac{2}{3}\Big(\frac{a^3+b^3}{a^2+b^2}\Big)\geq \frac{2}{3}\sqrt{ab}.$$ Thus $LHS \geq \dfrac{2}{3}(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq 2.$ 86. Romania 1997 $(a,b,c>0)$ $$\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\geq 1 \geq \frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}$$ By AM-GM inequality we have $\dfrac{a^2}{a^2+2bc}\geq \dfrac{a^2}{a^2+b^2+c^2}$ $\dfrac{b^2}{b^2+2ca}\geq \dfrac{b^2}{a^2+b^2+c^2}$ $\dfrac{c^2}{c^2+2ab}\geq \dfrac{c^2}{a^2+b^2+c^2}.$ Thus $LHS \geq 1$. The second inequality is equivalent to the first one by observing that $\dfrac{a^2}{a^2+2bc}= 1 - \frac{2bc}{a^2+2bc}$. 89. Estonia 1997 $(x,y\in\mathbb{R})$ $$x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$$ Let $a=\sqrt{x^2+1},b=\sqrt{y^2+1}$. Then by Cauchy-Schwarz inequality and AM-GM we get $$bx+ay\leq |bx+ay|\leq \sqrt{(a^2+b^2)(x^2+y^2)}\leq \frac{a^2+b^2+x^2+y^2}{2}=x^2+y^2+1.$$ 90. APMC 1996 $(x,y,z,t\in\mathbb{R},x+y+z+t=0,x^2+y^2+z^2+t^2=1)$ $$-1\leq xy+yz+zt+tx$$ By Cauchy-Schwarz inequality we have 91. Spain 1996 $(a,b,c>0)$$$|xy+yz+zt+tx|\leq x^2+y^2+z^2+t^2 = 1.$$ Thus $-1\leq xy+yz+zt+tx$. Since $(y+t)(x+z) =-(y+t)^2 \leq 0$, we get $xy+yz+zt+tx \leq 0.$ $$a^2+b^2+c^2-ab-bc-ca\geq 3(a-b)(b-c)$$ $$(a-2b+c)^2\geq 0.$$ |
96. Belarus 1996 $(x,y,z>0,x+y+z=\sqrt{xyz})$
$$xy+yz+zx\geq 9(x+y+z)$$ Note that $\sqrt{xyz} = x+y+z \geq 3\sqrt[3]{xyz}$ by AM-GM inequality. Thus $xyz\geq 3^6.$ Therefore, $$(xy+yz+zx)^2 \geq 3xyz(x+y+z) = 3\sqrt{xyz}(x+y+z)^2 \geq 81(x+y+z)^2.$$ 99. Baltic Way 1995 $(a,b,c,d>0)$ $$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$$ By AM-HM inequality, we have $\dfrac{1}{a+b}+\dfrac{1}{c+d}\geq \dfrac{4}{a+b+c+d}$ and $\dfrac{1}{b+c}+\dfrac{1}{d+a}\geq \dfrac{4}{a+b+c+d}$. Thus $$LHS = (a+c)\Big(\dfrac{1}{a+b}+\dfrac{1}{c+d}\Big)+(b+d)\Big(\frac{1}{b+c}+\frac{1}{d+a}\Big)\geq \frac{4(a+c)+4(b+d)}{a+b+c+d}=4.$$ 100. Canada 1995 $(a,b,c>0)$ $$a^ab^bc^c\geq (abc)^{\frac{a+b+c}{3}}$$ The inequality is equivalent to $$\Big(\frac{1}{a}\Big)^{\frac{a}{a+b+c}}\Big(\frac{1}{b}\Big)^{\frac{b}{a+b+c}}\Big(\frac{1}{c}\Big)^{\frac{c}{a+b+c}}\leq \frac{1}{\sqrt[3]{abc}}.$$ By Weighted AM-GM inequality we have $$LHS \leq \frac{1}{a}\cdot\frac{a}{a+b+c}+\frac{1}{b}\cdot\frac{b}{a+b+c}+\frac{1}{c}\cdot\frac{c}{a+b+c}= \frac{3}{a+b+c}\leq \frac{1}{\sqrt[3]{abc}}.$$ |
101. IMO 1995 $(abc=1,a,b,c>0)$
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq \frac{3}{2}$$ $LHS=\dfrac{b^2c^2}{ab+ca}+\dfrac{c^2a^2}{bc+ab}+\dfrac{a^2b^2}{ca+bc}$ $\geq \dfrac{(ab+bc+ca)^2}{2(ab+bc+ca)}$ $=\dfrac{ab+bc+ca}{2}$ $\geq \dfrac{3}{2}.$ 102. Russia 1995 $(x,y>0)$ $$\frac{1}{xy}\geq \frac{x}{x^4+y^2}+\frac{y}{y^4+x^2}$$ Note that $\dfrac{x}{x^4+y^2}\leq \frac{1}{2xy} \Leftrightarrow (x^2-y)^2\geq 0.$ Similarly, $\dfrac{y}{y^4+x^2}\leq \frac{1}{2xy} \Leftrightarrow (y^2-x)^2\geq 0.$ Thus $$RHS \leq \frac{1}{xy}.$$ 103 Macedonia1995 $(a,b,c>0)$ $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2$$ $\sqrt{\dfrac{a}{b+c}} = \dfrac{a}{\sqrt{a(b+c)}}\geq \dfrac{2a}{a+b+c}$ by AM-GM inequality. Similarly, $\sqrt{\dfrac{b}{c+a}} \geq \dfrac{2b}{a+b+c}$ and $\sqrt{\dfrac{c}{a+b}} \geq \dfrac{2c}{a+b+c}.$ Thus $$LHS \geq 2.$$ |
106. IMO Shortlist 1993 $(a,b,c,d>0)$
$$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\geq \frac{2}{3}$$ $LHS = \dfrac{a^2}{ab+2ac+3da}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{da+2bd+3cd}$ $\geq \dfrac{(a+b+c+d)^2}{4(ab+bc+cd+da+ac+bd)}$ $\geq \dfrac{2}{3}.$ The last inequality comes from $3(a+b+c+d)^2 \geq 8(ab+bc+cd+da+ac+bd)$ $\Leftrightarrow 3(a^2+b^2+c^2+d^2) \geq 2(ab+bc+cd+da+ac+bd)$ $\Leftrightarrow 2(a^2+b^2+c^2+d^2-ab-bc-cd-da) + (a-c)^2 + (b-d)^2 \geq 0.$ 108. Poland 1993 $(x,y,u,v>0)$ $$\frac{xy+xv+uy+uv}{x+y+u+v}\geq \frac{xy}{x+y}+\frac{uv}{u+v}$$ The inequality is equivalent to $$v^2x^2+u^2y^2 \geq 2uvxy$$ which is true by AM-GM inequality. 110. Italy 1993 $(0\leq a,b,c \leq 1)$ $$a^2+b^2+c^2 \leq a^2b+b^2c+c^2a + 1$$ We have $$1+ab+bc+ca\geq a+b+c \Leftrightarrow (1-a)(1-b)(1-c)+abc\geq 0.$$ Thus $$a^2(1-b)+b^2(1-c)+c^2(1-a) \leq a(1-b)+b(1-c)+c(1-a)\leq 1.$$ 115. IMO Shortlist 1990 $(ab+bc+cd+da=1,a,b,c,d>0)$ $$\frac{a^3}{b+c+d}+\frac{b^3}{c+d+a}+\frac{c^3}{d+a+b}+\frac{d^3}{a+b+c}\geq \frac{1}{3}$$ $LHS = \dfrac{a^4}{ab+ac+da}+\frac{b^4}{bc+bd+ab}+\frac{c^4}{cd+ac+bc}+\frac{d^4}{ad+bd+cd}$ $\geq \dfrac{(a^2+b^2+c^2+d^2)^2}{2(ab+bc+cd+da+ac+bd)}$ $\geq\dfrac{(ab+bc+cd+da)(a^2+b^2+c^2+d^2)}{2(ab+bc+cd+da+ac+bd)}$ $\geq \dfrac{1}{3}.$ The last inequality comes from $3(a^2+b^2+c^2+d^2)\geq 2(ab+bc+cd+da+ac+bd)$ $\Leftrightarrow 2(a^2+b^2+c^2+d^2-ab-bc-cd-da)+(a-c)^2+(b-d)^2\geq 0$ |
116. Lithuania 1987 $(x,y,z>0)$
$$\frac{x^3}{x^2+xy+y^2}+\frac{y^3}{y^2+yz+z^2}+\frac{z^3}{z^2+zx+x^2}\geq\frac{x+y+z}{3}$$ Note that $$\frac{x^3}{x^2+xy+y^2}\geq \frac{2x-y}{3}\Leftrightarrow (x+y)(x-y)^2\geq 0.$$ The above inequality also hold for the pairs $(y,z)$ and $(z,x)$. Adding the three inequalities we get the result. 126. Klamkin's inequality $(-1<x,y,z<1)$ $$\frac{1}{(1-x)(1-y)(1-z)}+\frac{1}{(1+x)(1+y)(1+z)}\geq 2$$ By AM-GM inequality, we have $LHS\geq \dfrac{2}{\sqrt{(1-x^2)(1-y^2)(1-z^2)}}\geq 2.$ 127. Carlson's inequality $(a,b,c>0)$ $$\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}\geq \sqrt{\frac{ab+bc+ca}{3}}$$ $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$ $=(a+b+c)(ab+bc+ca)-\sqrt[3]{ab\cdot bc\cdot ca}{\sqrt[3]{abc}}$ $\geq (a+b+c)(ab+bc+ca) - \dfrac{(a+b+c)(ab+bc+ca)}{9}$ $= \dfrac{8(a+b+c)(ab+bc+ca)}{9}$ $\geq 8\Big(\dfrac{ab+bc+ca}{3}\Big)^{3/2}.$ The last inequality comes from $3(ab+bc+ca)\leq (a+b+c)^2.$ 134. $abc=1,a,b,c>0$ $$\frac{1}{a^2(b+c)}+\frac{1}{b^2(c+a)}+\frac{1}{c^2(a+b)}\geq \frac{3}{2}$$ $$LHS =\frac{bc}{ab+ca}+\frac{ca}{bc+ab}+\frac{ab}{ca+bc}\geq\frac{3}{2}$$ by Nesbitt's inequality(#124). ขอจบเฉลยไว้แค่นี้นะครับ จริงๆยังมีข้อที่น่าสนใจอีกหลายข้อ แต่ส่วนใหญ่จะใช้แนวคิดเดิมๆ ผมข้ามโจทย์มาหลายข้อทีเดียวเพราะว่าใช้แนวคิดคล้ายๆกับข้อก่อนหน้านี้ หรือบางข้อจะเป็นโจทย์ระดับโอลิมปิกซึ่งสามารถหาเฉลยได้ง่ายครับก็เลยไม่เอามาลงไว้ แต่ถ้าใครอยากรู้ข้อไหนเพิ่มเติมก็ถามมาได้ครับ :cool: |
งั้นเด่วหลังผมกลับมาจาก สอวน. ที่ นครนายก เด่วจะมารวม ไฟล์ละกันนะคับ
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