NICE:easy and hard

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#1 h Jan 2, 2018, 2:27 AM • 4 Y

Y by rightways, Satops, Adventure10, Mango247

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

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#2 h Jan 2, 2018, 4:40 AM • 1 Y

Y by Adventure10

learningimprove wrote:

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

(2)Let $S_m=x_1^m+x_2^m+\cdots+x_n^m,x_1,x_2,\cdots,x_n\geq0,n,m\geq2,m,n\in{N},$ prove or disprove that
$S_m^m\leq S_{m+1}S_1^{m^2-m-1}.$ (It's a generalization of the famous inequalities for $m=2,3$ .)

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#3 h Jan 2, 2018, 6:13 AM • 1 Y

Y by Adventure10

(3)Let ${a,b,c}\geq0,m\geq{\frac{1}{\sqrt{2}}},$ then when $n=\frac{3m+2}{m}$ holds
$3abc+3m(n+a^2+b^2+c^2)\geq (1+3m+mn)(a+b+c)$

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#4 h Jan 2, 2018, 6:55 AM • 2 Y

Y by Adventure10, Mango247

learningimprove wrote:

(3)Let ${a,b,c}\geq0,m\geq{\frac{1}{\sqrt{2}}},$ then when $n=\frac{3m+2}{m}$ holds
$3abc+3m(n+a^2+b^2+c^2)\geq (1+3m+mn)(a+b+c)$

(4)Let ${a,b,c}\geq0,m\geq{\frac{1}{\sqrt{2}}},$ if
$3abc+3m(n+a^2+b^2+c^2)\geq (1+3m+mn)(a+b+c)$ holds then $n=\frac{3m+2}{m}.$

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#5 h Jan 2, 2018, 7:04 AM • 3 Y

Y by Adventure10, Mango247, rightways

learningimprove wrote:

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

By Holder $(a^8+b^8)(a+b)^{41}\geq\left(a^{\frac{49}{42}}+b^{\frac{49}{42}}\right)^{42}=\left(\left(a^{\frac{7}{6}}+b^{\frac{7}{6}}\right)^6\right)^7\geq(a^7+b^7)^7.$

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#6 h Jan 2, 2018, 7:05 AM • 2 Y

Y by Adventure10, Mango247

(5)Let $a,b,c>0,k\geq{\frac{1}{4}},$ prove that
$\sum_{cyc}{\frac{a}{\sqrt{b^2+c^2+kbc}}}\geq{\frac{3}{\sqrt{2+k}}}$ (6)Let $a,b,c>0,$ prove that
$\sum_{cyc}{\frac{a}{\sqrt{b^2+c^2+kbc}}}\geq{\frac{3}{\sqrt{2+k}}}$ holds if and only if it holds for $abc=0.$

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#7 h Jan 2, 2018, 7:07 AM • 2 Y

Y by Adventure10, Mango247

I think in this problem $abc=0$ is impossible.

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#8 h Jan 2, 2018, 7:37 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

I think in this problem $abc=0$ is impossible.

Should be changed to $abc\to{+0}.$ Thank You!

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#9 h Jan 2, 2018, 10:30 AM • 1 Y

Y by Adventure10

(7)Let $a+b+c\leq 2,a,b,c\geq0,$ prove that $\sum_{cyc}{a(1-b)(1-c)}\geq0$
(8)Let $a^m+b^m+c^m\leq 2{\cdot}k^m,a,b,c\geq0,k>0,m\neq0,$ prove or disprove that
$\sum_{cyc}{a^m(k-b)(k-c)}\geq0$

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#10 h Jan 2, 2018, 12:38 PM • 1 Y

Y by Adventure10

(9)Let $a,b,c>0,0<x\leq1,$ prove that
$\sum_{cyc}{(\frac{a}{b})^x}\geq \sum_{cyc}{(\frac{a+1}{b+1})^x}$
(10)Let $a,b,c,x,k>0,$ prove or disprove that
$\sum_{cyc}{(\frac{a}{b})^x}\geq \sum_{cyc}{(\frac{a+k}{b+k})^x}$

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#11 h Jan 2, 2018, 1:24 PM • 2 Y

Y by Adventure10, Mango247

learningimprove wrote:

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

(2)Let $S_m=x_1^m+x_2^m+\cdots+x_n^m,x_1,x_2,\cdots,x_n\geq0,n,m\geq2,m,n\in{N},$ prove or disprove that
$S_m^m\leq S_{m+1}S_1^{m^2-m-1}.$ (It's a generalization of the famous inequalities for $m=2,3$ .)

Using Hölder twice
$\begin{align*} S_{m+1}S_1^{m^2-m-1} &= (x_1^{m+1}+x_2^{m+1}+\cdots+x_n^{m+1})(x_1+x_2+\cdots+x_n)^{m^2-m-1}\\ &\ge (x_1+x_2+\cdots+x_n)^{m+1}(x_1+x_2+\cdots+x_n)^{m^2-m-1}\\ &=(x_1+x_2+\cdots+x_n)^{m^2}\ge(x_1^m+x_2^m+\cdots+x_n^m)^m\\ &=S_m^m \end{align*}$

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#12 h Jan 3, 2018, 12:37 AM • 2 Y

Y by Adventure10, Mango247

MaZDa wrote:

learningimprove wrote:

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

(2)Let $S_m=x_1^m+x_2^m+\cdots+x_n^m,x_1,x_2,\cdots,x_n\geq0,n,m\geq2,m,n\in{N},$ prove or disprove that
$S_m^m\leq S_{m+1}S_1^{m^2-m-1}.$ (It's a generalization of the famous inequalities for $m=2,3$ .)

Using Hölder twice
$\begin{align*} S_{m+1}S_1^{m^2-m-1} &= (x_1^{m+1}+x_2^{m+1}+\cdots+x_n^{m+1})(x_1+x_2+\cdots+x_n)^{m^2-m-1}\\ &\ge (x_1+x_2+\cdots+x_n)^{m+1}(x_1+x_2+\cdots+x_n)^{m^2-m-1}\\ &=(x_1+x_2+\cdots+x_n)^{m^2}\ge(x_1^m+x_2^m+\cdots+x_n^m)^m\\ &=S_m^m \end{align*}$

$(x_1^{m+1}+x_2^{m+1}+\cdots+x_n^{m+1})\geq (x_1+x_2+\cdots+x_n)^{m+1}?$

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#13 h Jan 3, 2018, 1:59 AM • 1 Y

Y by Adventure10

(11)Let $(a+b)(c+d)=2,a,b,c,d\geq0,$ prove that
$a^4+b^4+c^4+d^4\geq1$ (12)Let $(a+b)(c+d)=2k\geq0,a,b,c,d\geq0,$ prove that
$a^4+b^4+c^4+d^4\geq k^2$

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#14 h Jan 3, 2018, 2:04 AM • 2 Y

Y by Adventure10, Mango247

learningimprove wrote:

(7)Let $a+b+c\leq 2,a,b,c\geq0,$ prove that $\sum_{cyc}{a(1-b)(1-c)}\geq0$
(8)Let $a^m+b^m+c^m\leq 2{\cdot}k^m,a,b,c\geq0,k>0,m\neq0,$ prove or disprove that
$\sum_{cyc}{a^m(k-b)(k-c)}\geq0$

Seven is easy by uvw

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#15 h Jan 3, 2018, 2:10 AM • 2 Y

Y by Adventure10, Mango247

learningimprove wrote:

(11)Let $(a+b)(c+d)=2,a,b,c,d\geq0,$ prove that
$a^4+b^4+c^4+d^4\geq1$ (12)Let $(a+b)(c+d)=2k\geq1,a,b,c,d\geq0,$ prove or disprove that
$a^4+b^4+c^4+d^4\geq k^2$

11-12)

$8(a^4+b^4+c^4+d^4) \geq (a+b)^4+(c+d)^4 \geq 8k^2 \implies a^4+b^4+c^4+d^4 \geq k^2$

Where $8(a^4+b^4)\geq (a+b)^4$ by holder and then am-gm

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#86 h Jan 17, 2018, 1:01 PM • 1 Y

Y by Adventure10

arqady wrote:

learningimprove wrote:

(1)Let $a,b\geq 0,$ prove that $(a^7+b^7)^7\leq(a^8+b^8)(a+b)^{41}$

By Holder $(a^8+b^8)(a+b)^{41}\geq\left(a^{\frac{49}{42}}+b^{\frac{49}{42}}\right)^{42}=\left(\left(a^{\frac{7}{6}}+b^{\frac{7}{6}}\right)^6\right)^7\geq(a^7+b^7)^7.$

How $(a^{\frac{7}{6}}+b^{\frac{7}{6}})^6\ge (a^7+b^7)$ ?

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#87 h Jan 17, 2018, 1:17 PM • 1 Y

Y by Adventure10

learningimprove wrote:

Can u give solution for the (2) .Someone attempted it ,but i think that is wrong ,as he wrote that $S_{m+1}\ge (x_1+x_2+\cdots +x_n)^{m+1}$ ,which is wrong .

(2) is one of my guesses, and it hasn't been proven yet. $S_{m+1}\ge (x_1+x_2+\cdots +x_n)^{m+1}$ is wrong.[/quote]

As all $x_i\ge 0$ so by multinomial expansion it is false .

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#88 h Jan 17, 2018, 9:43 PM • 2 Y

Y by targo___, Adventure10

targo___ wrote:

[

How $(a^{\frac{7}{6}}+b^{\frac{7}{6}})^6\ge (a^7+b^7)$ ?

$(a^{\frac{7}{6}}+b^{\frac{7}{6}})^6=x^7+6\left(a^{\frac{7}{6}}\right)^5b^{\frac{7}{6}}+...+b^7.$

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#89 h Jan 18, 2018, 1:31 AM • 1 Y

Y by Adventure10

learningimprove wrote:

(73)Let $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c,a,b,c>0,$ prove that
$\prod_{cyc}{(a^{18}-a^{17}+1)}\geq{\prod_{cyc}{a^{18}}-\prod_{cyc}{a^{17}}+1}$

(74)Let $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=a_1+a_2+\cdots+a_n,a_1,a_2,\cdots,a_n>0,$

$p>q,{p,q}\in\mathbb{N_+},$ prove or disprove that
$\prod_{i=1}^n{(a_i^{p}-a_i^{q}+1)}\geq{\prod_{i=1}^n{a_i^{p}}-\prod_{i=1}^n{a_i^{q}}+1}$

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#90 h Jan 18, 2018, 6:12 AM • 3 Y

Y by maXplanK, Adventure10, Mango247

Just genaralize #5 to get the result
$(x_1^{m+1}+x_2^{m+1}+\cdots +x_n^{m+1})^{\frac{1}{m^2-m}}(x_1+x_2+\cdots +x_n)^{\frac{m^2-m-1}{m^2-m}}$ $\ge x_1^{\frac{m+1}{m^2-m}+\frac{m^2-m-1}{m^2-m}}+x_2^{\frac{m^2}{m^2-m}}+\cdots +x_n^{\frac{m2}{m^2-m}}$ $~~=(x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})$ Now, $S_{m+1}S_1^{m^2-m-1}\ge (x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})^{m^2-m}=((x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})^{m-1})^m$
Now we need $(x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})\ge (x_1^m+x_2^m+\cdots +x_n^m)^{\frac{1}{m-1}}$
which is true by multinomial expansion ( $x_i\ge 0 ,\forall i\in \{1,2,\cdots ,n\}$ )
Hence, $S_{m+1}S_1^{m^2-m-1}\ge (x_1^m+x_2^m+\cdots +x_n^m)^m=S_m^m$

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#91 h Jan 18, 2018, 7:16 AM • 2 Y

Y by Adventure10, Mango247

targo___ wrote:

Just genaralize #5 to get the result
$(x_1^{m+1}+x_2^{m+1}+\cdots +x_n^{m+1})^{\frac{1}{m^2-m}}(x_1+x_2+\cdots +x_n)^{\frac{m^2-m-1}{m^2-m}}$ $\ge x_1^{\frac{m+1}{m^2-m}+\frac{m^2-m-1}{m^2-m}}+x_2^{\frac{m^2}{m^2-m}}+\cdots +x_n^{\frac{m2}{m^2-m}}$ $~~=(x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})$ Now, $S_{m+1}S_1^{m^2-m-1}\ge (x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})^{m^2-m}=((x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})^{m-1})^m$
Now we need $(x_1^{\frac{m}{m-1}}+x_2^{\frac{m}{m-1}}+\cdots +x_n^{\frac{m}{m-1}})\ge (x_1^m+x_2^m+\cdots +x_n^m)^{\frac{1}{m-1}}$
which is true by multinomial expansion ( $x_i\ge 0 ,\forall i\in \{1,2,\cdots ,n\}$ )
Hence, $S_{m+1}S_1^{m^2-m-1}\ge (x_1^m+x_2^m+\cdots +x_n^m)^m=S_m^m$

Wow, just to genaralize the idea of #5, great!

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#92 h Jan 18, 2018, 11:07 AM • 1 Y

Y by Adventure10

(75)Let $a+b+c=1,a,b,c\geq0,$ prove that
$\prod_{cyc}{(a^{20}-a^{18}+1)}\leq{\prod_{cyc}{a^{20}}-\prod_{cyc}{a^{18}}+1}$

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#93 h Jan 18, 2018, 10:55 PM • 2 Y

Y by Adventure10, Mango247

(76)Let $a_1+a_2+\cdots+a_n\leq1,a_1,a_2,\cdots,a_n>0,p>q,{p,q}\in\mathbb{N_+},$ prove or disprove that
$\prod_{i=1}^n{(a_i^{p}-a_i^{q}+1)}\leq{\prod_{i=1}^n{a_i^{p}}-\prod_{i=1}^n{a_i^{q}}+1}$
If let $a_1+a_2+\cdots+a_n=n,$ prove or disprove that above inequality retrorse holds.

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#95 h Jan 19, 2018, 8:03 AM • 2 Y

Y by Adventure10, Mango247

(77)Let $a^2+b^2+c^2={d^2+e^2},a^3+b^3+c^3\geq{d^3+e^3},a,b,c,d,e\geq0,$ prove that
$a^4+b^4+c^4\geq{d^4+e^4}$

(78)Let $a^2+b^2+c^2={d^2+e^2},a^3+b^3+c^3\geq{d^3+e^3},a,b,c,d,e\geq0,n\geq4,$

$n\in\mathbb{N},,$ prove or disprove that
$a^n+b^n+c^n\geq{d^n+e^n}$

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#96 h Jan 19, 2018, 8:12 AM • 2 Y

Y by Adventure10, Mango247

What is the point of posting so many inequalities on the same thread when you leave no time for anyone to reply?!

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#97 h Mar 1, 2018, 10:13 AM • 1 Y

Y by Adventure10

Thanks for so many inequalities ,however,can you give the answer to above problems?

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#98 h Jul 5, 2023, 3:20 AM

Y by

learningimprove wrote:

(21)Let $a,b,c\geq0,$ prove that
$(a^2+1)(b^2+1)(c^2+1)\geq(a+b+c-1)^2+2(a-1)(b-1)(c-1)$ (22)Let $a,b,c,d\geq0,$ prove that
$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\geq(a+b+c+d-1)^2$ $(a^2+1)(b^2+1)(c^2+1)(d^2+1)\geq(a+b+c+d-1)^2-2(a-1)(b-1)(c-1)(d-1)$

Let $a,b,c\geq0,$ prove that
$(a^2+1)(b^2+1)(c^2+1)\geq(a+b+c-1)^2$

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#99 h Jul 5, 2023, 3:26 AM

Y by

learningimprove wrote:

(19)Let $a,b,c\geq0,$ prove that
$(a^2+b^2+\frac{1}{4})(b^2+c^2+\frac{1}{4})(c^2+a^2+\frac{1}{4})\geq(2a-\frac{1}{4})(2b-\frac{1}{4})(2c-\frac{1}{4})$ (20)Let $a,b,c\geq0,$ find all $k\in\mathbb{R},$ such that holds
$(a^2+b^2+\frac{1}{2}-k)(b^2+c^2+\frac{1}{2}-k)(c^2+a^2\frac{1}{2}-k)\geq(2a-k)(2b-k)(2c-k)$

Let $a,b,c$ be positive real numbers .Prove that $( a^2+b+\dfrac{3}{4})(b^2+c+\dfrac{3}{4})(c^2+a+\dfrac{3}{4})\geq (2a+\dfrac{1}{2})(2b+\dfrac{1}{2})(2c+\dfrac{1}{2})$

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#100 h Apr 17, 2025, 2:32 AM

Y by

learningimprove wrote:

(30)Let $a+b+c+d=0,a^2+b^2+c^2+d^2=4,a,b,c,d\in\mathbb{R},$ prove that
$abcd\leq1$

Let $a,b,c,d$ be reals such that $a+b+c+d =0$ and $a^2+b^2+c^2+d^2=4.$ Prove that $-\frac 13\leq abcd\leq 1$ lbh_qys

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#101 h Apr 17, 2025, 2:53 AM

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Let $a,b,c,d>0$ and $\frac 1a + \frac 1b + \frac 1c + \frac 1d =1.$ Prove that $(a^2-4a+\frac 92)(b^2-4b+ \frac 92) (c^2-4c+\frac 92) (d^2-4d+\frac 92) \geq \frac{6561}{16}$ $(a^2-4a+\frac{21}{5})(b^2-4b+ \frac{21}{5}) (c^2-4c+\frac{21}{5}) (d^2-4d+\frac{21}{5}) \geq \frac{194481}{625}$

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