MathDB
Prove this inequality of radii of cicumsphere and insphere of a tetrahedron

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
In the any [ABCD][ABCD] tetrahedron we denote with α\alpha, β\beta, γ\gamma the measures, in radians, of the angles of the three pairs of opposite edges and with rr, RR the lengths of the rays of the sphere inscribed and respectively circumscribed the tetrahedron. Demonstrate inequality(3rR)3≤sin⁡α+β+γ3\left(\frac{3r}{R}\right)^3\leq \sin \frac{\alpha +\beta +\gamma}{3}(A refinement of inequality R≥3rR \geq 3r).
tetrahedron3D geometrycircumradiusinradiusinequalitiesgeometry
Prove that the sequence is convergent and find its limit

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let be x1=1n!n+1x_1=\frac{1}{\sqrt[n+1]{n!}} and x2=1(n−1)!n+1x_2=\frac{1}{\sqrt[n+1]{(n-1)!}} for all n∈N∗n\in \mathbb{N}^* and f:(1(n+1)!n+1,1]→Rf:\left(\left .\frac{1}{\sqrt[n+1]{(n+1)!}},1\right.\right] \to \mathbb{R} where f(x)=n+1xln⁡(n+1)!+(n+1)ln⁡(xx)f(x)=\frac{n+1}{x\ln (n+1)!+(n+1)\ln \left(x^x\right)}Prove that the sequence (an)n≥1(a_n)_{n\geq1} when an=∫x1x2f(x)dxa_n=\int \limits_{x_1}^{x_2}f(x)dx is convergent and compute lim⁡n→∞an\lim \limits_{n \to \infty}a_n
Sequenceslimitintegrationcalculus
Prove this 5 variable inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let aa, bb, cc, dd, ee be real strictly positive real numbers such that abcde=1abcde = 1. Then is true the following inequality:dea(b+1)+eab(c+1)+abc(d+1)+bcd(e+1)+cde(a+1)≥52\frac{de}{a(b+1)}+\frac{ea}{b(c+1)}+\frac{ab}{c(d+1)}+\frac{bc}{d(e+1)}+\frac{cd}{e(a+1)}\geq \frac{5}{2}
inequalities
Prove this 8 variable inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
If aa, bb, c≥1c \geq 1; y≥x≥1y \geq x \geq 1; pp, qq, r>0r > 0 then(1+y(apbqcr)1p+q+r1+x(apbqcr)1p+q+r)p+q+r(apbqcr)1p+q+r(1+xa1+ya)pa(1+xb1+yb)qb(1+xc1+yc)rc\left(\frac{1+y\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}{1+x\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}\right)^{\frac{p+q+r}{\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}}\left(\frac{1+xa}{1+ya}\right)^{\frac{p}{a}}\left(\frac{1+xb}{1+yb}\right)^{\frac{q}{b}}\left(\frac{1+xc}{1+yc}\right)^{\frac{r}{c}} ≥∏cyc(1+y(apbq)1p+q1+x(apbq)1p+q)p+q(apbq)1p+q\geq \prod \limits_{cyc}\left(\frac{1+y\left(a^pb^q\right)^{\frac{1}{p+q}}}{1+x\left(a^pb^q\right)^{\frac{1}{p+q}}}\right)^{\frac{p+q}{\left(a^pb^q\right)^{\frac{1}{p+q}}}}
inequalities
Prove this inequality on radius of spheres of a tetrahedron

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
In the [ABCD][ABCD] tetrahedron having all the faces acute angled triangles, is denoted by rXr_X, RXR_X the radius lengths of the circle inscribed and circumscribed respectively on the face opposite to the X∈{A,B,C,D}X \in \{A,B,C,D\} peak, and with RR the length of the radius of the sphere circumscribed to the tetrahedron. Show that inequality occurs8R2≥(rA+RA)2+(rB+RB)2+(rC+RC)2+(rD+RD)28R^2 \geq (r_A + R_A)^2 + (r_B + R_B)^2 + (r_C + R_C)^2 + (r_D + R_D)^2
tetrahedroncircumsphereinsphere3D geometryinequalitiesexsphere
Prove this 3 real number variable inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
If aa, bb, c∈Rc \in \mathbb{R} then∑cyc(c+a)2b2+c2a2+5∣∑cycab∣≥∑cyc(ab+2bc+ca)2+(b+c)2a2\sum \limits_{cyc} \sqrt{(c+a)^2b^2+c^2a^2}+\sqrt{5}\left |\sum \limits_{cyc} \sqrt{ab}\right |\geq \sum \limits_{cyc}\sqrt{(ab+2bc+ca)^2+(b+c)^2a^2}
Summationinequalities
Prove this algebraic inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let a1,a2,⋯ ,ana_1,a_2,\cdots ,a_n be nn positive numbers such that ∑i=1nai=n\sum \limits_{i=1}^n\sqrt{a_i}=\sqrt{n}. Then∏i=1n−1(1+1ai)ai+1(1+1an)a1≥1+n∑i=1nai\prod \limits_{i=1}^{n-1}\left(1+\frac{1}{a_i}\right)^{a_{i+1}}\left(1+\frac{1}{a_n}\right)^{a_1}\geq 1+\frac{n}{\sum \limits_{i=1}^na_i}
ProductSummationinequalities
Find the limit of this summation

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Compute lim⁡n→∞1n∑k=1nn+1n+k+1−nn+kn+1n+k−nn+k\lim \limits_{n \to \infty}\frac{1}{n}\sum \limits_{k=1}^n\frac{\sqrt[n+k+1]{n+1}-\sqrt[n+k]{n}}{\sqrt[n+k]{n+1}-\sqrt[n+k]{n}}
limitSummationcalculus
Prove this integral inequality involving pi

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Prove that ∫0π2(cos⁡x)1+2n+1dx≤2n−1n!π2(2n+1)!\int \limits_0^{\frac{\pi}{2}}(\cos x)^{1+\sqrt{2n+1}}dx\leq \frac{2^{n-1}n!\sqrt{\pi}}{\sqrt{2(2n+1)!}}for all n∈N∗n\in \mathbb{N}^*
integrationinequalitiescalculus
Prove this 6 variable inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
If bk≥ak≥0b_k \geq a_k \geq 0 (k=1,2,3)(k = 1, 2, 3) and α≥1\alpha \geq 1 then(α+3)∑cyc(b1−a1)((b2+b3)α+2+(a2+a3)α+2−(a2+b3)α+1−(b2+a3)α+1)(\alpha+3)\sum \limits_{cyc}(b_1-a_1)\left((b_2+b_3)^{\alpha+2}+(a_2+a_3)^{\alpha+2}-(a_2+b_3)^{\alpha+1}-(b_2+a_3)^{\alpha+1}\right) ≤(α+2)(α+3)∑cyc(b1−a1)(b2−a2)(b3α+1−a3α+1)\leq (\alpha+2)(\alpha+3)\sum \limits_{cyc}(b_1-a_1)(b_2-a_2)(b_3^{\alpha+1}-a_3^{\alpha+1}) +(b3+b2+a1)α+3+(b3+a2+a1)α+3+(a3+b2+a1)α+3+(a3+a2+b1)α+3+ (b_3 + b_2 + a_1)^{\alpha+3}+(b_3 + a_2 + a_1)^{\alpha+3}+(a_3 + b_2 + a_1)^{\alpha+3}+(a_3 + a_2 + b_1)^{\alpha+3} −(b3+b2+b1)α+3−(b3+a2+a1)α+3−(a3+b2+b1)α+3−(a3+a2+a1)α+3-(b_3 + b_2 + b_1)^{\alpha+3}-(b_3 + a_2 + a_1)^{\alpha+3}-(a_3 + b_2 + b_1)^{\alpha+3}-(a_3 + a_2 + a_1)^{\alpha+3}
Summationinequalities
Prove this inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Let x1,x2,≥,xnx_1, x_2,\geq , x_n be a positive numbers, k≥1k \geq 1. Then the following inequality is true: (x1k+x2k+⋯+xnk)k+1≥(x1k+1+x2k+1⋯+xnk+1)k+2(∑1≤i<j≤nxikxj)k\left(x_1^k+x_2^k+\cdots +x_n^k\right)^{k+1}\geq \left(x_1^{k+1}+x_2^{k+1}\cdots +x_n^{k+1}\right)^k+2\left(\sum \limits_{1\leq i<j\leq n}x_i^kx_j\right)^k
inequalitiesSummation
Prove these two inequalities of tetrahedron

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
In all tetrahedron ABCDABCD holds
[*] ∑cycha−rha+r≥∑cychat−rt(ha+r)t\displaystyle{\sum \limits_{cyc}\frac{h_a-r}{h_a+r}\geq \sum \limits_{cyc}\frac{h_a^t-r^t}{(h_a+r)^t}} [*] ∑cyc2ra−r2ra+r≥∑cyc2rat−rt(2ra+r)t\displaystyle{\sum \limits_{cyc}\frac{2r_a-r}{2r_a+r}\geq \sum \limits_{cyc}\frac{2r_a^t-r^t}{(2r_a+r)^t}}
for all t∈[0,1]t\in [0,1]
tetrahedron3D geometryinequalitiesexradiusinradius
Solve for x, y, z, t

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Prove that exist different natural numbers xx, yy, zz, tt for which 256×2019180n+1=2x9−2y6+z5−t4256\times 2019^{180n+1}=2x^9-2y^6+z^5-t^4for all n∈N∗n\in \mathbb{N}^*
number theory
Prove this inequality as the bound of the integration

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
If 0<a≤b0 < a \leq b then23tan⁡−1(2(b2−a2)(a2+2)(b2+2))≤∫ab(x2+1)(x2+x+1)(x3+x2+1)(x3+x+1)dx≤43tan⁡−1((b−a)3a+b+2(1+ab))\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2(b^2 - a^2)}{(a^2+2)(b^2+2)}\right)\leq \int \limits_a^b \frac{(x^2+1)(x^2+x+1)}{(x^3 + x^2 + 1) (x^3 + x + 1)}dx\leq \frac{4}{\sqrt{3}}\tan^{-1}\left(\frac{(b-a)\sqrt{3}}{a+b+2(1+ab)}\right)
integrationinequalitiestrigonometryinverse trigonometric functioncalculus
Prove this geometric inequality for external and internal bisector

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Denote wa‾,wb‾,wc‾\overline{w_a}, \overline{w_b}, \overline{w_c} the external angle-bisectors in triangle ABCABC, prove that ∑cyc1wa≤(s2−r2−4Rr)(8R2−s2−r2−2Rr)8s2R2r\sum \limits_{cyc} \frac{1}{w_a}\leq \sqrt{\frac{(s^2 - r^2 - 4Rr)(8R^2 - s^2 - r^2 - 2Rr)}{8s^2R^2r}}
geometryTriangleangle bisectorgeometric inequalityinequalitiescircumradiusinradius
Prove this trigonometric inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
If x∈(0,π2)x \in \left(0,\frac{\pi}{2}\right) then(sin⁡(π2sin⁡x)sin⁡x)2+(sin⁡(π2cos⁡x)cos⁡x)2≥3\left(\frac{\sin \left(\frac{\pi}{2}\sin x\right)}{\sin x}\right)^2+\left(\frac{\sin \left(\frac{\pi}{2}\cos x\right)}{\cos x}\right)^2\geq 3
trigonometryinequalities
Prove this geometric inequality

Source: 2019 Jozsef Wildt International Math Competition

5/20/2020
Denote TT the Toricelli point of the triangle ABCABC. Prove that AB2×BC2×CA2≥3(TA2×TB+TB2×TC+TC2×TA)(TA×TB2+TB×TC2+TC×TA2)AB^2 \times BC^2 \times CA^2 \geq 3(TA^2\times TB + TB^2 \times TC + TC^2 \times TA)(TA\times TB^2 + TB \times TC^2 + TC \times TA^2)
geometrytoricelli pointgeometric inequalityinequalities
Miklos Schweitzer 1949_2

Source:

10/2/2008
Compute \lim_{n\rightarrow \infty} \int_{0}^{\pi} \frac {\sin{x}}{1 \plus{} \cos^2 nx}dx .
limitintegrationtrigonometryreal analysisreal analysis unsolved
Miklos Schweitzer 1949_1

Source:

10/2/2008
Let an infinite sequence of measurable sets be given on the interval (0,1) (0,1) the measures of which are ≥α>0 \geq \alpha>0. Show that there exists a point of (0,1) (0,1) which belongs to infinitely many terms of the sequence.
limitreal analysisreal analysis unsolved
Miklos Schweitzer 1949_5

Source:

10/2/2008
Let f(x) f(x) be a polynomial of second degree the roots of which are contained in the interval [\minus{}1,\plus{}1] and let there be a point x_0\in [\minus{}1.\plus{}1] such that |f(x_0)|\equal{}1. Prove that for every α∈[0,1] \alpha \in [0,1], there exists a \zeta \in [\minus{}1,\plus{}1] such that |f'(\zeta)|\equal{}\alpha and that this statement is not true if α>1 \alpha>1.
algebrapolynomialalgebra proposed
Miklos Schweitzer 1949_3

Source:

10/2/2008
Let p p be an odd prime number and a1,a2,...,ap a_1,a_2,...,a_p and b1,b2,...,bp b_1,b_2,...,b_p two arbitrary permutations of the numbers 1,2,...,p 1,2,...,p . Show that the least positive residues modulo p p of the numbers a1b1,a2b2,...,apbp a_1b_1, a_2b_2,...,a_pb_p never form a permutation of the numbers 1,2,...,p 1,2,...,p.
number theory proposednumber theory
Miklos Schweitzer 1949_7

Source:

10/2/2008
Find the complex numbers z z for which the series 1 \plus{} \frac {z}{2!} \plus{} \frac {z(z \plus{} 1)}{3!} \plus{} \frac {z(z \plus{} 1)(z \plus{} 2)}{4!} \plus{} \cdots \plus{} \frac {z(z \plus{} 1)\cdots(z \plus{} n)}{(n \plus{} 2)!} \plus{} \cdots converges and find its sum.
Gaussfunctionintegrationtrigonometrycomplex numberscomplex analysiscomplex analysis unsolved
Miklos Schweitzer 1949_8

Source:

10/2/2008
The four sides of a skew quadrangle and the two segments joining the midpoints of the opposite sides are realized by rigid bars. The bars are linked by hinges. Prove that this apparatus is not rigid. P.S: The 1949 Miklos Schweitzer competition had only 8 problems!
geometry
Miklos Schweitzer 1949_6

Source:

10/2/2008
Let n n and k k be positive integers, n≥k n\geq k. Prove that the greatest common divisor of the numbers \binom{n}{k},\binom{n\plus{}1}{k},\ldots,\binom{n\plus{}k}{k} is 1 1.
number theorygreatest common divisornumber theory proposed
Miklos Schweitzer 1949_4

Source:

10/2/2008
Let A A and B B be two disjoint sets in the interval (0,1) (0,1) . Denoting by μ \mu the Lebesgue measure on the real line, let μ(A)>0 \mu(A)>0 and μ(B)>0 \mu(B)>0 . Let further n n be a positive integer and \lambda \equal{}\frac1n . Show that there exists a subinterval (c,d) (c,d) of (0,1) (0,1) for which \mu(A\cap (c,d))\equal{}\lambda \mu(A) and \mu(B\cap (c,d))\equal{}\lambda \mu(B) . Show further that this is not true if λ \lambda is not of the form 1n \frac1n.
real analysisreal analysis unsolvedcollege contestsMiklos Schweitzer