All questions with tag [math: a.m.-g.m.-inequality]


If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$

If $a,b,c,d$ declare integers and also $a+b+c+d=16$, confirm that $$\left(a+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(b+\frac{1}{d}\right)^2+\left(d+\frac{1}{b}\right)^2 \geq \frac{289}{4}.$$ I recognize I need to make use of some inequality, not exactly sure AM GM will certainly function below or Minkowski inequality. Yet I jus...
2022-07-12 12:01:50

Find the minimum value of the quantity where $a , b , c$ are real positive numbers.

Find the minimum value of the amount where $a , b , c$ are actual favorable numbers. $$\left(\frac{a^2 + 3a + 1}{a}\right) \left(\frac{b^2 +3b + 1}{b}\right)\left(\frac{c^2 + 3c + 1}{c}\right) $$ I assume the to get the solution we require to make use of $A.M.\ge G.M.$ How i can attain this?
2022-07-11 20:39:49

Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$

Let $a,b,c,d,e$ declare actual numbers which please $abcde=1$. Just how can one confirm that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$
2022-06-09 11:51:48

Combined AM GM QM inequality

I found this intriguing inequality, and also was seeking intriguing evidence. $x,y,z \geq 0$. $$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$ Addendum. As a whole, when is $$ a\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+b\sqrt [3]{xyz}\leq (a+b)\left(\frac{x+y+z}{3}\right) $$ real?
2022-06-08 21:41:36

Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

take into consideration favorable numbers $a_1,a_2,a_3,\ldots,a_n$ and also $b_1,b_2,\ldots,b_n$. does the adhering to in - equal rights holds and also if it does after that just how to confirm it $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}$
2022-06-06 10:51:56