Questions tagged [a.m.-g.m.-inequality]

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

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6 votes
3 answers
168 views

Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum?

Does the following function have a global minimum: $$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$$ where $x \in \mathbb{N}$? I tried using WolframAlpha, but it appears to give an inconsistent result....
3 votes
6 answers
389 views

Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$

prove the following inequality: $$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers. Thanks :)
  • 6,670
3 votes
2 answers
938 views

Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$

In the pdf which you can download here I found the following inequality which I can't solve it. Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{...
  • 6,670
6 votes
2 answers
2k views

Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$

Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$ Prove the inequality $$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$ Thanks :)
  • 6,670
2 votes
4 answers
1k views

Minimum value of given expression

What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$ I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can ...
8 votes
4 answers
483 views

Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$

Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? : $$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
  • 2,680
12 votes
5 answers
581 views

proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$.

Let $a,b,c>0$ how to prove that : $$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$ I find that $$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\...
  • 2,680
3 votes
2 answers
245 views

Maximum product $ab$ for $a+b=k$ - proof based on AM-GM inequality

My question is: Prove- If $a,b$ are two positive real numbers such that their sum is $a+b=k$. Then the product $ab$ is maximum if and only if $a=b=\displaystyle\frac{k}{2}$. I proved the ...
  • 955
8 votes
4 answers
294 views

Theorem about two real numbers

My question is: $a,b$ are two positive real numbers such that their product is constant,equal to $k$ say. Prove: the sum $a+b$ is minimum if and only if $a = b= \sqrt k$. Can this be solved using $A....
  • 955
8 votes
2 answers
290 views

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 $a,b,c,d$ are positive integers and $a+b+c+d=16$, prove 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}{...
2 votes
1 answer
159 views

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

Find the minimum value of the quantity where $a , b , c$ are real positive 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 ...
  • 2,591
8 votes
3 answers
1k views

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 $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove 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{{...
15 votes
2 answers
1k views

Combined AM GM QM inequality

I came across this interesting inequality, and was looking for interesting proofs. $x,y,z \geq 0$ $$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$ Addendum....
  • 4,858
14 votes
3 answers
1k views

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

consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{...
  • 671

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