解答下載: 2010年數學奧林匹克不等式解答

問題1.(Albania BMO TST)設[latex]a,b,c[/latex]為某三角形的三邊長,對於不等式

[latex]a^3+b^3+c^3<k(a+b+c)(ab+bc+ca)[/latex]

(a)證明[latex]k=1[/latex]時不等式成立.
(b)求使得不等式恒成立的最小常數[latex]k[/latex].

問題2.(Greece)已知[latex]x,y>0,x+y=2a[/latex],求證:[latex]x^3y^3(x^2+y^2)^2\le 4a^{10}.[/latex]

問題3. (India)已知[latex]a,b,c>0[/latex]且[latex]ab+bc+ca\le 3abc[/latex].求證:

[latex]\displaystyle \sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+\sqrt{\frac{a^2+b^2}{a+b}}+3\le \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}).[/latex]

問題4. (India)設[latex]P[/latex]是三角形[latex]ABC[/latex]的Brocard點,求證:

[latex] \displaystyle \left(\frac{AP}{BC}\right)^2+\left(\frac{BP}{CA}\right)^2+\left(\frac{CP}{AB}\right)^2\ge 1.[/latex]

問題5. (Iran)已知[latex]a,b,c>0[/latex],求證:

[latex]\displaystyle \frac{1}{a^2}+\frac{1}{b^2}+ \frac{1}{c^2}+ \frac{1}{(a+b+c)^2}\ge \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^2.[/latex]

問題6. (Iran)已知[latex]x,y,z>0,xy+yz+zx=1[/latex],求證:

[latex]\displaystyle 3-\sqrt{3}+ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge (x+y+z)^2.[/latex]

問題7. (Korea)任意給定一個三角形[latex]ABC,P,Q,R[/latex]是三角形[latex]ABC[/latex]內切圓分別在邊 [latex]BC,CA,AB[/latex]的切點.設[latex]T[/latex]是三角形[latex]ABC[/latex]的面積,[latex]L[/latex]是其外接圓直徑,求證:

[latex]\displaystyle \left(\frac{AB}{PQ}\right)^{3}+\left(\frac{BC}{QR}\right)^{3}+\left(\frac{CA}{RP}\right)^{3}\ge\frac{2}{\sqrt{3}}\frac{L^{2}}{T}. [/latex]

問題8.(Portugal)證明:對任意的三角形都存在某兩邊長[latex]a,b[/latex],滿足

[latex]\displaystyle \frac{\sqrt{5}-1}{2}<\frac{a}{b}<\frac{\sqrt{5}+1}{2}.[/latex]

問題9. (Vietnam)已知[latex]\displaystyle a,b,c>0,16(a+b+c)\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/latex].求證:

[latex]\displaystyle \sum_{cyc}\left(\frac{1} {a+b+ \sqrt{2a+2c}}\right)^3\le \frac{8}{9}.[/latex]

問題10. (Indonesia)已知[latex]a,b,c\ge 0[/latex]和[latex]x,y,z>0[/latex]且[latex]a+b+c=x+y+z[/latex],求證:

[latex]\displaystyle \frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c.[/latex]

問題11. (Indonesia)設[latex]a_1,a_2,\cdots[/latex]為實數列,滿足[latex]\displaystyle a_1=1,a_2=\frac{4} {3},a_{n+1}= \sqrt{1+a_na_{n-1}},\forall n\ge 2[/latex].求證:對任意的[latex]\displaystyle n\ge 2,a_n^2>a_{n-1}^2+\frac{1}{2}[/latex]並且

[latex]\displaystyle 1+\frac{1}{a_1}+\frac{1} {a_2}+ \cdots+\frac{1}{a_n}>2a_n.[/latex]

問題12.(Indonesia)實係數多項式[latex]f(x)=ax^3+bx^2+cx+d[/latex]有3個正實根,且[latex]f(0)<0[/latex].求證:

[latex]2b^3+9a^2d-7abc\le 0.[/latex]

問題13. (Kazakhstan)實數[latex]A[/latex]給定,求最大的實數[latex]M[/latex]使得

[latex]\displaystyle\frac{1} {x}+\frac{1}{y}+\frac{A}{x+y}\ge \frac{M}{\sqrt{xy}}[/latex]

對[latex]x,y>0[/latex]恒成立.

問題14.(Kazakhstan)已知[latex]x,y\ge 0[/latex],求證:

[latex]\sqrt{x^2-x+1}\sqrt{y^2-y+1}+\sqrt{x^2+x+1}\sqrt{y^2+y+1}\ge 2(x+y).[/latex]

問題15. (Switzerland)已知[latex]x,y,z>0,xyz=1[/latex],求證:

[latex]\displaystyle \frac{(x+y-1)^2} {z}+\frac{(y+z-1)^2}{x}+\frac{(z+x-1)^2}{y}\ge x+y+z.[/latex]

問題16. (Turkey)已知[latex]a,b,c>0[/latex],求證:

[latex]\displaystyle \sum_{cyc}\sqrt[4]{\frac{(a^2+b^2)(a^2- ab+b^2)}{2}}\le \frac{2}{3}(a^2+b^2+c^2)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right).[/latex]

問題17.(USA TST)正實數[latex]a,b,c[/latex]滿足[latex]abc=1[/latex].求證:

[latex]\displaystyle\frac{1} {a^5(b+2c)^2}+\frac{1}{b^5(c+2a)^2}+\frac{1}{c^5(a+2b)^2}\ge \frac{1}{3}.[/latex]

問題18.(USA TST)[latex]P[/latex]為三角形[latex]ABC[/latex]內一點,求證:

[latex]\displaystyle\frac{AP} {h_b+h_c}+\frac{BP}{h_c+h_a}+\frac{CP}{h_a+h_b}\ge 1.[/latex]

問題19.(Balkan)已知[latex]a,b,c>0[/latex],求證:

[latex]\displaystyle\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a}\ge 0.[/latex]

問題20.(Middle European)對任意給定的整數[latex]n\ge 2[/latex],求最大的[latex]C_n[/latex]使得對任意的正實數[latex]a_1,a_2,\cdots,a_n[/latex]都有

[latex]\displaystyle\frac{a_1^2+\cdots+a_n^2}{n} \ge\left (\frac{a_1+\cdots+a_n}{n}\right)^2+C_n(a_1-a_n)^2.[/latex]

問題21.(Mediterranean )已知[latex]n>2[/latex],正實數[latex]a_1,a_2,\cdots,a_n[/latex]滿足 [latex]a_1+a_2+\cdots+a_n=1[/latex],求 證:

[latex]\displaystyle\frac{a_{2}a_{3}\cdots a_{n}}{a_{1}+n-2}+\frac{a_{1}a_{3}\cdots a_{n}}{a_{2}+n-2}+\cdots+\frac{a_{1}a_{2}\cdots a_{n-1}}{a_{n}+n-2}\le\frac{1}{(n-1)^{2}} ,[/latex]

並求等號成立條件.

問題22.(北 方)設正實數[latex]a,b,c[/latex]滿足[latex](a+2b)(b+2c)=9[/latex].求證:

[latex]\displaystyle\sqrt{\frac{a^2+b^2}{2}}+2\sqrt[3]{\frac{b^3+c^3}{2}}\ge 3.[/latex]

問題23.(北方)設[latex]\displaystyle x,y,z\in [0,1],|x-y|\le\frac{1}{2},|z-y|\le\frac{1}{2},|x-z|\le\frac{1}{2} [/latex],求

[latex]W=x+y+z-xy-yz-zx[/latex]

的最大值和最小值.

問題24.(東南)設 [latex]n[/latex]為正整數,實數[latex]a_1,a_2,\cdots,a_n,r_1,r_2,\cdots, r_n[/latex]滿足[latex]a_1\le a_2\le \cdots\le a_n[/latex]和[latex]0\le r_1\le r_2\le \cdots\le r_n[/latex],求證:

[latex]\displaystyle\sum_{i=1}^n\sum_{j=1}^n a_ia_j \min(r_i,r_j)\ge 0.[/latex]

問題25.(女子)設實數 [latex]x_1,x_2,\cdots,x_n[/latex]滿足[latex]\displaystyle\sum_{i=1}^n x_i^2=1,n\ge 2[/latex],令[latex]\displaystyle S=\sum_{i=1}^nix_i^2[/latex].求證:

[latex]\displaystyle\sum_{k=1}^n\left(1- \frac{k}{S}\right)^2\frac{x_k^2} {k}\le \left(\frac{n-1}{n+1}\right)^2\sum_{k=1}^n \frac{x_k^2}{k},[/latex]

並求等號成立條件.

問題26.(西部)設非負實數[latex]a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n[/latex]同時滿足如下條件:
(1)[latex]\displaystyle\sum_{i=1}^{n}(a_i+b_i)=1[/latex]; (2)[latex]\displaystyle\sum_{i=1}^{n}i(a_i-b_i)=0[/latex]; (3)[latex]\displaystyle\sum_{i=1}^{n}i^2(a_i+b_i)=10[/latex].
求證:[latex]\displaystyle\max\{a_k,b_k\}\le \frac{10}{10+k^2}[/latex]對任意的[latex]1\le k \le n[/latex]成立.

27. (聯賽)給定正整數[latex]n> 2[/latex],設正實數[latex]a_1,a_2,\cdots,a_n[/latex]滿足[latex]a_k\le 1,k=1,2,\cdots,n[/latex].記[latex]\displaystyle A_k=\frac{a_1+\cdots+a_k}{k}[/latex], [latex]k=1,\cdots,n[/latex].求證:

[latex]\displaystyle\left| \sum_{i=1}^na_i-\sum_{i=1}^nA_i\right|<\frac{n-1}{2}.[/latex]

問題28. (Turkey)正實數[latex]a_1,a_2,\cdots,a_n[/latex]滿足[latex]a_1a_2 \cdots a_n=1[/latex],求證:

[latex] \displaystyle\sum_{i=1}^{n}\frac{a_i}{\sqrt{a_i^4+3}}\le \sum_{i=1}^{n}\frac{1}{2a_i}. [/latex]

問題29.(Bulgaria)設正實數[latex]a,b,c[/latex]滿足[latex]a+b+c=3[/latex],求證:

[latex]abc(a^2+b^2+c^2)\le 3[/latex]

問題30.(集訓隊)給定正整數[latex]n\ge 2[/latex]和正實數[latex]a[/latex],正實數[latex]x_1,x_2,\cdots,x_n[/latex]满足[latex]\displaystyle\prod\limits_{i=1}^n x_i=1[/latex]. 求最小的实数 [latex]M=M(n,a)[/latex] , 使得

[latex]\displaystyle\sum_{i=1}^n \frac{1}{a+S-x_i}\le M[/latex]

恒成立 , 其中[latex]S=x_1+x_2+\cdots+x_n[/latex] .