Articles of a.m. g.m. inequality

Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$

Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers). There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!

Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$

Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$ Attempt The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$ I don’t […]

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $a+b+{1\over{ab}}$

If $a^2+b^2=1$, where $a>0$ and $b>0$, then find the minimum value of $a+b+{1\over{ab}}$ This can be easily done by calculas but is there any way to do do this by algebra

Wanted: Low-dimensional SOS certificate for the AM-GM inequality

Consider the AM-GM inequality in five variables $$a^5+b^5+c^5+d^5+e^5-5abcde\;\ge 0\qquad\forall\,a,b,c,d,e\ge 0\,.$$ Can one write the LHS as a concrete (finite) sum $\,\sum_i h_i\,s_i\,$ with real polynomials $\,h_i(a,b,c,d,e)\,$ and $\,s_i(a,b,c,d,e)$, where each $\,h_i\,$ is positive and homogeneous of degree $1$, and each $\,s_i\,$ is a square? For the AM-GM inequality in $3$ variables the answer would be […]

SOS: Proof of the AM-GM inequality

“A basic strategy in tackling inequalities of few variables is to write things into sum of squares…” is a quote from an answer, intended as commenting the post entitled “Can this inequality proof be demystified?”. I’d like to know what the Sum Of Squares strategy yields for the AM-GM inequality with $n$ variables. In the […]

If $a+b+c=6$ and $a,b,c$ belongs to positive reals $\mathbb{R}^+$; then find the minimum value of $\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$ .

If $a+b+c=6$ and $a,b,c$ belongs to positive reals, then find the minimum value of $$\frac{1}{a}+\frac{4}{b}+\frac{9}{c}$$ using AM $\ge HM$ $\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ ${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge\frac{3}{2}$ or **why not $AM\ge GM $ $\frac{\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+a+b+c}{6}\ge (\frac{1}{a}\times\frac{4}{b}\times\frac{9}{c}\times a\times b\times c)^\frac{1}{6} $ $\Rightarrow \frac{1}{a}+\frac{4}{b}+\frac{9}{c}\ge 6(6^\frac{1}{3}-1)$**

Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.

Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$. How to prove inequality $$ ab^2+bc^2+ca^2\le 4.\tag{*} $$ In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality $$ 27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**} $$ $\color{gray}{\mbox{(Without using “universal” Lagrange multipliers method).}}$ Thanks!

Solving Equation through inequalities.

If $x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0$ has positive roots then find $a,b,c,d$. I did something but that don’t deserve to be added here, but what I thought before doing that is following: For us, Product and Sum of roots are given. Roots are positive. Hence I tried to use AM-GM-HM inequalities, as sum and product are given, but I […]

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. In general, when is $$ a\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+b\sqrt [3]{xyz}\leq (a+b)\left(\frac{x+y+z}{3}\right) $$ true?

Is this continuous analogue to the AM–GM inequality true?

First let us remind ourselves of the statement of the AM–GM inequality: Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $N\geqslant 1$ non-negative real numbers, we have $$\frac1N\sum_k x_k \geqslant \left(\prod_k x_k\right)^{\frac1N}$$ It is well known that the sum operator ∑ can be generalised so that it operates on continuous functions rather than on discrete […]