2010年数学奥林匹克不等式问题 2012年数学奥林匹克不等式问题 2011年数学奥林匹克不等式问题 2010年数学奥林匹克不等式问题 Algebraic Inequalities I , Vasile Cirtoaje Algebraic Inequalities II , Vasile Cirtoaje
問題1.(Albania BMO TST) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1818014&sid=9ae4875dcdd8df450f0f1e77759e5453#p1818014 設[latex]a,b,c[/latex]為某三角形的三邊長,對於不等式\[a^3+b^3+c^3 (b)求使得不等式恒成立的最小常數[latex]k[/latex]. 回复 ↓
問題2.(Greece) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1790813&sid=9ae4875dcdd8df450f0f1e77759e5453#p1790813 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] 回复 ↓
問題3. (India) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1882531&sid=9ae4875dcdd8df450f0f1e77759e5453#p1882531 已知[latex]a,b,c>0[/latex]且[latex]ab+bc+ca\le 3abc[/latex].求證:\[\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+\sqrt{\frac{a^2+b^2}{a+b}}+3\le \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}).\] 回复 ↓
問題4. (India) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1883908&sid=9ae4875dcdd8df450f0f1e77759e5453#p1883908 設[latex]P[/latex]是三角形[latex]ABC[/latex]的Brocard點,求證:\[\left(\frac{AP}{BC}\right)^2+\left(\frac{BP}{CA}\right)^2+\left(\frac{CP}{AB}\right)^2\ge 1.\] 回复 ↓
問題5. (Iran) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1974763&sid=9ae4875dcdd8df450f0f1e77759e5453#p1974763 已知[latex]a,b,c>0[/latex],求證:\[\frac{1}{a^2}+\frac{1}{b^2}+ \frac{1}{c^2}+ \frac{1}{(a+b+c)^2}\ge \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^2.\] 回复 ↓
問題6. (Iran) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1974770&sid=9ae4875dcdd8df450f0f1e77759e5453#p1974770 已知[latex]x,y,z>0,xy+yz+zx=1[/latex],求證:\[3-\sqrt{3}+ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge (x+y+z)^2.\] 回复 ↓
問題7. (Korea) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1826698&sid=9ae4875dcdd8df450f0f1e77759e5453#p1826698 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:\[\left(\frac{AB}{PQ}\right)^{3}+\left(\frac{BC}{QR}\right)^{3}+\left(\frac{CA}{RP}\right)^{3}\ge\frac{2}{\sqrt{3}}\frac{L^{2}}{T}. \] 回复 ↓
問題8.(Portugal) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000026&sid=9ae4875dcdd8df450f0f1e77759e5453#p2000026 證明:對任意的三角形都存在某兩邊長[latex]a,b[/latex],滿足\[\frac{\sqrt{5}-1}{2}<\frac{a}{b}<\frac{\sqrt{5}+1}{2}.\] 回复 ↓
問題9. (Vietnam) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2061442&sid=9ae4875dcdd8df450f0f1e77759e5453#p2061442 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:\[\sum_{cyc}\left(\frac{1} {a+b+ \sqrt{2a+2c}}\right)^3\le \frac{8}{9}.\] 回复 ↓
問題10. (Indonesia) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1679875&sid=9ae4875dcdd8df450f0f1e77759e5453#p1679875 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:\[\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c.\] 回复 ↓
問題1.(Albania BMO TST) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1818014&sid=9ae4875dcdd8df450f0f1e77759e5453#p1818014
設[latex]a,b,c[/latex]為某三角形的三邊長,對於不等式\[a^3+b^3+c^3
問題2.(Greece) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1790813&sid=9ae4875dcdd8df450f0f1e77759e5453#p1790813
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) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1882531&sid=9ae4875dcdd8df450f0f1e77759e5453#p1882531
已知[latex]a,b,c>0[/latex]且[latex]ab+bc+ca\le 3abc[/latex].求證:\[\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+\sqrt{\frac{a^2+b^2}{a+b}}+3\le \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}).\]
問題4. (India) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1883908&sid=9ae4875dcdd8df450f0f1e77759e5453#p1883908
設[latex]P[/latex]是三角形[latex]ABC[/latex]的Brocard點,求證:\[\left(\frac{AP}{BC}\right)^2+\left(\frac{BP}{CA}\right)^2+\left(\frac{CP}{AB}\right)^2\ge 1.\]
問題5. (Iran) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1974763&sid=9ae4875dcdd8df450f0f1e77759e5453#p1974763
已知[latex]a,b,c>0[/latex],求證:\[\frac{1}{a^2}+\frac{1}{b^2}+ \frac{1}{c^2}+ \frac{1}{(a+b+c)^2}\ge \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b+c}\right)^2.\]
問題6. (Iran) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1974770&sid=9ae4875dcdd8df450f0f1e77759e5453#p1974770
已知[latex]x,y,z>0,xy+yz+zx=1[/latex],求證:\[3-\sqrt{3}+ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\ge (x+y+z)^2.\]
問題7. (Korea) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1826698&sid=9ae4875dcdd8df450f0f1e77759e5453#p1826698
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:\[\left(\frac{AB}{PQ}\right)^{3}+\left(\frac{BC}{QR}\right)^{3}+\left(\frac{CA}{RP}\right)^{3}\ge\frac{2}{\sqrt{3}}\frac{L^{2}}{T}. \]
問題8.(Portugal) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2000026&sid=9ae4875dcdd8df450f0f1e77759e5453#p2000026
證明:對任意的三角形都存在某兩邊長[latex]a,b[/latex],滿足\[\frac{\sqrt{5}-1}{2}<\frac{a}{b}<\frac{\sqrt{5}+1}{2}.\]
問題9. (Vietnam) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2061442&sid=9ae4875dcdd8df450f0f1e77759e5453#p2061442
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:\[\sum_{cyc}\left(\frac{1} {a+b+ \sqrt{2a+2c}}\right)^3\le \frac{8}{9}.\]
問題10. (Indonesia) http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1679875&sid=9ae4875dcdd8df450f0f1e77759e5453#p1679875
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:\[\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\ge a+b+c.\]