# Questions tagged [geometric-inequalities]

This is a tag for geometric problems involving inequalities.

429 questions
Filter by
Sorted by
Tagged with
108 views

• 1,954
31 views

### A Pedoe-like Geometric Inequality

Problem Given two triangles $ABC$ and $A'B'C'$ where $a,b,c$ and $a',b',c'$ are the corresponding sides and $F, F'$ denotes the areas of the two triangles. Prove: a^2a'^2+b^2b'^2+c^2c'...
• 549
36 views

57 views

### Sum of distances from interior point to vertices is greater than double the sum of distances to edges

Let $\triangle ABC$, $M\in Int(\triangle ABC)$. Let $MD\perp BC$, $ME\perp AC$ and $MF\perp AB$, $D\in BC$, $E\in AC$, $F\in AB$. Prove that $$MA+MB+MC\geq 2\cdot(MD+ME+MF).$$ My only idea is to use ...
• 867
23 views

### Lower bound on inradius in terms of diameter of isotropic convex body

Consider a convex body whose tensor of inertia is a multiple of the identity matrix. This means that the body is in some appropriate sense balanced or isotropic. Can one lower bound the inradius (...
• 1
1 vote
94 views

• 139
102 views

### Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk

In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$ Here ...
• 16.6k
76 views

### $(2a^2+bc)(2b^2+ca)(2c^2+ab)\ge(2a^2+2b^2-c^2)(2b^2+2c^2-a^2)(2c^2+2a^2-b^2)$

Problem: Prove that in any triangle ABC, the following inequality is true: $$(2a^2+bc)(2b^2+ca)(2c^2+ab)\ge(2a^2+2b^2-c^2)(2b^2+2c^2-a^2)(2c^2+2a^2-b^2)$$ Anyone can help me? I tried AM-GM for right ...
• 405
115 views

### Prove that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$ if $a^2+b^2+c^2+abc=4$.

Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4$. How do you prove that $\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$? My Approach: I ...
137 views

### In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$

Problem: Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$ Happy Vietnamese Women's ...
• 405
111 views

• 103
123 views

• 1,185
49 views

### If $H$ is the orthocenter and $D,E,F$ are foot of altitudes prove that $\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$

Let ABC be a triangle with $H$ as its the orthocenter and $AD,BE,CF$ as the three altitudes. Prove that $$\frac{AD}{HD}+\frac{BC}{HE}+\frac{CF}{HF}≥9$$ I substituted $AD$ for $AH+HD$, changing the ...
• 1,185
145 views

### If $2$ altitudes of a triangle are $9$ and $40$ then find the minimum perimeter

If $2$ altitudes of a triangle with integer side lengths are $9$ and $40$ units in length, then find the minimum possible perimeter of the triangle Since the altitude is the shortest distance from a ...
• 1,185
1 vote
61 views

### proving $\min(OR,PR,QR)+OR+PR+QR<OQ+QP+PO$

In the figure prove that $$\min(OR,PR,QR)+OR+PR+QR<OQ+QP+PO$$ This is a stronger form of the inequality $OR+PR+QR<OQ+QP+PO$ that can be easily proven.I have made some constructions in the ...
• 11.7k
41 views

### Equivalence of statements via geometric inequality on $4$ Points

Let us consider $4$ Points $A,B,C,D$ on a line $g$. We are to prove: If $\vert AP\vert+\vert PD\vert >\vert BP\vert+\vert CP\vert$ for all $P$ not on $g$, then $\vert AB\vert=\vert CD\vert$. I ...
• 548
1 vote
114 views

### Prove this geometric inequality

Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$ My attempt: I think ...
• 81
These trangles have the same perimeter 2q，in which condition，can we have the trangle with the maximum area? I have tried to use the Heron's formula. $S=\sqrt{q(q-a)(q-b)(q-c)}$ I have sought the ...