Let $a,b,c\ge0$ and $a+b+c\ge3.$ Prove that $a^4+b^3+c^2\ge a^3+b^2+c.$
Problem
Source: Kyiv mathematical festival 2019
Tags: Kyiv mathematical festival, inequalities, BPSQ
10.05.2019 03:25
rogue wrote: Let $a,b,c\ge0$ and $a+b+c\ge3.$ Prove that $a^4+b^3+c^2\ge a^3+b^2+c.$ Proof of Zhangyunhua:
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10.05.2019 03:41
Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $ a_1+a_2+\cdots+a_n\geq n.$ Prove that $$a_1^{n+1}+a_2^n+\cdots+a_{n-1}^3+a_n^2\geq a_1^{n}+a_2^{n-1}+\cdots+a_{n-1}^2+a_n.$$
11.05.2019 01:37
rogue wrote: Let $a,b,c\ge0$ and $a+b+c\ge3.$ Prove that $a^4+b^3+c^2\ge a^3+b^2+c.$ Solution. By the Cauchy-Schwarz's Inequality we have $$a^4+b^3+c^2=\frac{a^6}{a^2}+\frac{b^4}{b}+\frac{c^2}{1}\ge\frac{\left(a^3+b^2+c\right)^2}{a^2+b+1}.$$Hence it remains to show $a^3+b^2+c\ge a^2+b+1$. Indeed, from the assumption $a+b+c\ge3$ it follows that \begin{align*}&a^3+b^2+c-\left(a^2+b+1\right)\\ \ge&a^3+b^2+(3-a-b)-\left(a^2+b+1\right)\\ =&a^3-a^2-a+1+b^2-2b+1\\ =&(a-1)^2(a+1)+(b-1)^2\ge0, \end{align*}which establishes the result. $\blacksquare$
11.05.2019 02:00
sqing wrote: Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $ a_1+a_2+\cdots+a_n\geq n.$ Prove that $$a_1^{n+1}+a_2^n+\cdots+a_{n-1}^3+a_n^2\geq a_1^{n}+a_2^{n-1}+\cdots+a_{n-1}^2+a_n.$$ Solution. Since $x^{k+1}-x^k-x+1=\left(x^k-1\right)(x-1)=\left(x^{k-1}+x^{k-2}+\cdots+1\right)(x-1)^2\ge0$ for any $x>0$ and positive integer $k$, we have $x^{k+1}-x^k\ge x-1$ for any $x>0$ and positive integer $k$, which implies that \begin{align*}&a_1^{n+1}+a_2^n+\cdots+a_{n-1}^3+a_n^2-\left(a_1^{n}+a_2^{n-1}+\cdots+a_{n-1}^2+a_n\right)\\ \ge&(a_1-1)+(a_2-1)+\cdots+(a_n-1)\\ =&a_1+a_2+\cdots+a_n-n\ge0. \end{align*}As required. $\blacksquare$
11.05.2019 03:07
Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be non-negative numbers such that $a_1^2+a_2^2+\cdots+a_n^2=n.$ Prove that $$a_1+a_2+\cdots+a_n \geq n-2+2\sqrt{a_1a_2\cdots a_n}.$$
28.03.2020 02:32
rogue wrote: Let $a,b,c\ge0$ and $a+b+c\ge3.$ Prove that $$a^4+b^3+c^2\ge a^3+b^2+c.$$ Let a,b,c be positive real numbers such that $abc\ge 1. $ Prove that$$ a^4+b^3+c^2\ge a^3+b^2+c$$
29.03.2020 08:10
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