Question

Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$

score 230 · Answer 1 · 2023-04-02 06:22:25

230

Views

Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$

Published on 02 Apr 2023 - 6:22

score 53 · Answer 2 · 2023-04-17 10:58:26

53

Views

Is the geometric mean of two non-negative, twice-differentiable concave real functions concave?

Published on 17 Apr 2023 - 10:58

score 138 · Answer 3 · 2026-03-29 06:02:29

138

Views

Do the $4$ Pythagorean means divide into two algebraic dualities?

Published on 29 Mar 2026 - 6:02

score 111 · Answer 4 · 2023-04-22 23:20:33

111

Views

Prove that $\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$ if $x$ and $y$ are real positive numbers!

Published on 22 Apr 2023 - 23:20

score 153 · Answer 5 · 2023-04-23 11:55:28

153

Views

Let $a, b, c$ be positive real numbers. Prove the inequality.

Published on 23 Apr 2023 - 11:55

score 72 · Answer 6 · 2026-03-26 04:20:02

72

Views

Proof of inequality $x + x^{-1} - x^r - x^{-r} \geq 0$ for $x>0$ and $0<r<1$.

Published on 26 Mar 2026 - 4:20

score 241 · Answer 7 · 2023-05-12 12:02:53

241

Views

Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$

Published on 12 May 2023 - 12:02

score 3.3k · Answer 8 · 2026-03-25 20:35:58

3.3k

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How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$

Published on 25 Mar 2026 - 20:35

score 28 · Answer 9 · 2023-05-16 17:27:15

28

Views

Prove $x^2+4y^2<1$ given $x^3 + y^3 = x-y$ and x and y are positive real numbers. (Using AM-GM)

Published on 16 May 2023 - 17:27

score 49 · Answer 10 · 2026-03-25 19:10:01

49

Views

Determine the minimal value of $\frac{3a}{b+c}+\frac{4b}{c+a}+\frac{5c}{a+b}$

Published on 25 Mar 2026 - 19:10

score 139 · Answer 11 · 2023-05-18 12:39:46

139

Views

Show that $\frac{x^2}{x+1} + \frac{(1-x)^2}{2-x} \geq \frac 13$ for $0\leq x \leq 1$

Published on 18 May 2023 - 12:39

score 116 · Answer 12 · 2023-05-18 20:08:10

116

Views

Equality in the A.M G.M inequality

Published on 18 May 2023 - 20:08

score 42 · Answer 13 · 2026-03-26 06:09:09

42

Views

bounding $(|\alpha|+1)^2+(x+|\alpha|)^2$

Published on 26 Mar 2026 - 6:09

score 178 · Answer 14 · 2023-05-19 22:19:32

178

Views

AM-GM Inequality. Let a, b, c be positive real numbers. Prove that $ \frac {a+b+c}{3} \cdot (a^2+b^2+c^2) \ge a^2b + b^2c + c^2a$.

Published on 19 May 2023 - 22:19

score 751 · Answer 15 · 2026-03-25 23:42:37

751

Views

If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$

Published on 25 Mar 2026 - 23:42

List Question

Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$

Is the geometric mean of two non-negative, twice-differentiable concave real functions concave?

Do the $4$ Pythagorean means divide into two algebraic dualities?

Prove that $\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$ if $x$ and $y$ are real positive numbers!

Let $a, b, c$ be positive real numbers. Prove the inequality.

Proof of inequality $x + x^{-1} - x^r - x^{-r} \geq 0$ for $x>0$ and $0<r<1$.

Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$

How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$

Prove $x^2+4y^2<1$ given $x^3 + y^3 = x-y$ and x and y are positive real numbers. (Using AM-GM)

Determine the minimal value of $\frac{3a}{b+c}+\frac{4b}{c+a}+\frac{5c}{a+b}$

Show that $\frac{x^2}{x+1} + \frac{(1-x)^2}{2-x} \geq \frac 13$ for $0\leq x \leq 1$

Equality in the A.M G.M inequality

bounding $(|\alpha|+1)^2+(x+|\alpha|)^2$

AM-GM Inequality. Let a, b, c be positive real numbers. Prove that $ \frac {a+b+c}{3} \cdot (a^2+b^2+c^2) \ge a^2b + b^2c + c^2a$.

If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$

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