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inequalities
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sqing   29
N 14 minutes ago by sqing
Let $ a,b,c\geq 0 $ and $ a^2+ab+bc+ca=3 .$ Prove that$$\frac{1}{1+a^2}+ \frac{1}{1+b^2}+ \frac{1}{1+c^2} \geq \frac{3}{2}$$$$\frac{1}{1+a^2}+ \frac{1}{1+b^2}+ \frac{1}{1+c^2}-bc \geq -\frac{3}{2}$$
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sqing
Jul 19, 2024
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inequality
revol_ufiaw   3
N Apr 5, 2025 by MS_asdfgzxcvb
Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
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revol_ufiaw
Apr 5, 2025
MS_asdfgzxcvb
Apr 5, 2025
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revol_ufiaw
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#1 h  Apr 5, 2025, 2:05 PM
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Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
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sqing
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sqing
#2 h  Apr 5, 2025, 2:18 PM • 1 Y
Y by revol_ufiaw
revol_ufiaw wrote:
Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
Prove that :$\forall a,b \in \mathbb{R}^+$
$$ a^b(b+1)^{b+1}\le b^b (a+1)^{b+1}$$
This post has been edited 1 time. Last edited by sqing, Apr 5, 2025, 2:18 PM
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revol_ufiaw
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#3 h  Apr 5, 2025, 2:31 PM
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sqing wrote:
revol_ufiaw wrote:
Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
Prove that :$\forall a,b \in \mathbb{R}^+$
$$ a^b(b+1)^{b+1}\le b^b (a+1)^{b+1}$$

Thanks. Can the inequality be proven without derivatives?
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MS_asdfgzxcvb
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MS_asdfgzxcvb
#4 h  Apr 5, 2025, 2:55 PM • 2 Y
Y by revol_ufiaw, KevinKV01
sqing wrote:
revol_ufiaw wrote:
Prove that that for any real $x \ge 0$ and natural number $n$,
$$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$$
Prove that :$\forall a,b \in \mathbb{R}^+$
$$ a^b(b+1)^{b+1}\le b^b (a+1)^{b+1}$$
For \(a\neq b\), Bernoulli's Inequality, \(y:=\frac{b+1}{a-b}>\frac b{a-b}:=x\color{blue}\xRightarrow{\hspace{35pt}}\left(1+\frac 1y\right)^{y/x}\ge\left(1+\frac 1x\right).\)
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