The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that: $$\frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2} \leq 1$$ When do we have equality in the right inequality and when in the left inequality? Proposed by Walther Janous
Problem
Source: 48th Austrian Mathematical Olympiad Beginners' Competition(2017)
Tags: inequalities, algebra, Austria
28.02.2018 03:18
The RHS is $a^3+b^3 \leq a^2+b^2$, which is clear as $a,b \leq 1$. The LHS is $a^2 + (1-a)^2 \leq 2a^3 + 2(1-a)^3$, which is equivalent to $2a^2 -2a +1 \leq 6a^2-6a+2$, or $(2a-1)^2 \geq 0$, trivially true.
28.02.2018 03:19
Medjl wrote: The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that: $$\frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2} \leq 1$$ When do we have equality in the right inequality and when in the left inequality? Proposed by Walther Janous $$a^3+a^3+b^3\ge 3a^2b,a^3+b^3+b^3\ge 3ab^2\implies a^3+b^3\geq a^2b+ab^2 \iff \frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2}$$
28.02.2018 03:34
Medjl wrote: The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that: $$\frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2} \leq 1$$ When do we have equality in the right inequality and when in the left inequality? Proposed by Walther Janous Hello ,Walther Janous!
28.02.2018 03:39
you wrote this in two posts ? weird ?
28.02.2018 04:17
Medjl wrote: you wrote this in two posts ? weird ? Sorry. Walther Janous is my idol of worship. Thanks. Proof of Zhangyunhua:
Attachments:

28.02.2018 12:07
Medjl wrote: The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that: $$\frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2} \leq 1$$ When do we have equality in the right inequality and when in the left inequality? Proposed by Walther Janous $\frac{1}{2}\leq \frac{a^3+b^3}{a^2+b^2}\leq 1$ $\iff$ $\frac{1}{2}(a^3+a^2b+ab^2+b^3)\leq a^3+b^3\leq a^3+a^2b+ab^2+b^3$
28.02.2018 16:59
\begin{align*} a^2-2ab+b^2=(a-b)^2&\ge0 \\ \frac{ab}{a^2+b^2}&\le\frac12\\ \frac{ab}{a^2+b^2}-(a+b)^2\le\frac{1}2-1&=-\frac12 \\ \frac{-a^2-ab-b^2}{a^2+b^2}&\le -\frac12 \\ \frac{a^3+b^3}{a^2+b^2}&\ge\frac{1}2\end{align*}
04.03.2018 03:44
Medjl wrote: The nonnegative real numbers $a$ and $b$ satisfy $a + b = 1$. Prove that: $$\frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2} \leq 1$$ When do we have equality in the right inequality and when in the left inequality? Proposed by Walther Janous $a^3+b^3=(a+b)(a^2-ab+b^2)=a^2-ab+b^2.$ $\therefore \frac{1}{2} \leq \frac{a^3+b^3}{a^2+b^2}\leq 1$ $\Longleftrightarrow \frac 12\leq 1-\frac{ab}{a^2+b^2}\leq 1$ $\Longleftrightarrow 0\leq \frac{1}{a^2+b^2}\leq \frac 12$ $\Longleftrightarrow a^2+b^2\geq 2ab$ $\Longleftrightarrow (a-b)^2\geq 0.$ Equality hods, in the right inequality, $a+b=1$ and $ab=0\Longleftrightarrow (a,\ b)=(1,\ 0),\ (0,\ 1)$ which satisfy $a,\ b\geq 0$, and in the left inequality, $a+b=1$ and $a^2+b^2=2ab\Longleftrightarrow a=b=\frac 12$ which satisfy $a,\ b\geq 0.$
04.03.2018 04:44
a^2 + (1-a)^2 = 2a^2 - 2a + 1 a^3 + (1-a)^3 = 1 - 3a + 3a^2 3a^2 - 3a + 1 2a^2 - 2a + 1 3a(a - 1) + 1 2a(a - 1) + 1 Since 0 ≤ a ≤ 1the top it less than the bottom thus the ≤ 1 portion is satisfied. To prove the other requirement, we have 2(3a(a-1) + 1) ≥ 2a(a - 1) + 1 -> 6a(a-1) + 1 ≥ 2a(a-1) -> 4a(a-1) + 1 ≥ 0. The vertex of this parabola has a y-value of 1 - (-4)^2/4 = 0. Since it opens upwards and this is a legal value of a, the proof is complete.
04.03.2018 04:45
For equality for the right and left just plug in the same way I did with b = 1 - a
11.01.2021 18:10
Pretty trivial we see that for proving lhs our. Inequality transforms to $ab\ leq\1/4$ which is am gm on a+b=1 now for rhs it transforms. To $ab\geq\0 $ which is. True since a and b are nonnegative
03.11.2021 04:34
Let $a$ and $b$ be nonnegative numbers such that $a+b=1$. Prove that $$\frac{3-\sqrt 3}{4}\leq \frac{a^3+b^3}{a^2+2b^2}\leq 1$$$$\frac{4-\sqrt{7}}{6}\leq \frac{a^3+b^3}{a^2+3b^2}\leq 1$$$$\frac{k+1-\sqrt{k^2-k+1}}{2k}\leq \frac{a^3+b^3}{a^2+kb^2}\leq 1$$Where $k\in N^+.$ $$\frac{15-3\sqrt 6-\sqrt{6\sqrt{6}-9}}{8}\leq \frac{a^3+2b^3}{a^2+b^2}\leq 2$$$$\frac{2}{3}\leq \frac{a^{3} + 4b^{3}}{a^{2} +2 b^{2}} \leq 2$$ Generalization of Austrian 2017
Attachments:

06.09.2023 14:20
First, we prove, that $1 \ge \frac{a^3+b^3}{a^2+b^2}=1$, so $a^2+b^2 \ge a^3+b^3$, note that $a \le 1, b\le 1$, hence, $a^2 \ge a^3, b^2 \ge b^3$, so we have the desired. Inequality, holds, when $(a,b)=(1,0), (0,1).$ oops why cant i prove the LHS