Prove the sum is greater than or equal to one half

[anemone]

Let $a,\,b$ and $c$ be positive real numbers for which $a+ b + c = 1$.

Prove that \(\displaystyle \frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+a^2}+\frac{c^3}{a^2+b^2}\ge \frac{1}{2}.\)

 

[MarkFL]

My solution:

Spoiler

Let:

\(\displaystyle f(a,b,c)=\frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+a^2}+\frac{c^3}{a^2+b^2}\)

By cyclic symmetry, we see that the extremum occurs at:

\(\displaystyle (a,b,c)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)\)

And we find:

\(\displaystyle f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\frac{1}{2}\)

Choosing another point on the constraint, such as:

\(\displaystyle (a,b,c)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{2}\right)\)

We find:

\(\displaystyle f\left(\frac{1}{4},\frac{1}{4},\frac{1}{2}\right)=\frac{3}{4}>\frac{1}{2}\)

Hence, we may conclude:

\(\displaystyle f_{\min}=\frac{1}{2}\)

 

[anemone]

My solution:

Spoiler

Let:

\(\displaystyle f(a,b,c)=\frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+a^2}+\frac{c^3}{a^2+b^2}\)

By cyclic symmetry, we see that the extremum occurs at:

\(\displaystyle (a,b,c)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)\)

And we find:

\(\displaystyle f\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=\frac{1}{2}\)

Choosing another point on the constraint, such as:

\(\displaystyle (a,b,c)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{2}\right)\)

We find:

\(\displaystyle f\left(\frac{1}{4},\frac{1}{4},\frac{1}{2}\right)=\frac{3}{4}>\frac{1}{2}\)

Hence, we may conclude:

\(\displaystyle f_{\min}=\frac{1}{2}\)

Very good, MarkFL!(Cool) And thanks for participating!

 

[lfdahl]

My solution:

Spoiler

WLOG we can let $a \ge b \ge c$, and we get:

\[\frac{1}{b^2+c^2} \geq \frac{1}{a^2+c^2}\geq \frac{1}{a^2+b^2}\]Repeated use of Chebyschevs Sum Inequality:\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq 9\underbrace{(a+b+c)}_{=1} \left ( \frac{a^2}{b^2+c^2} + \frac{b^2}{a^2+c^2} + \frac{c^2}{a^2+b^2} \right )\\\\ 9 \left ( \frac{a^2}{b^2+c^2} + \frac{b^2}{a^2+c^2} + \frac{c^2}{a^2+b^2} \right )\geq 3\underbrace{(a+b+c)}_{=1} \left ( \frac{a}{b^2+c^2} + \frac{b}{a^2+c^2} + \frac{c}{a^2+b^2} \right ) \\\\ 3 \left ( \frac{a}{b^2+c^2} + \frac{b}{a^2+c^2} + \frac{c}{a^2+b^2} \right )\geq \underbrace{(a+b+c)}_{=1} \left ( \frac{1}{b^2+c^2} + \frac{1}{a^2+c^2} + \frac{1}{a^2+b^2} \right )\]Using the Arithmetic Harmonic Mean Inequality:\[\frac{1}{b^2+c^2} + \frac{1}{a^2+c^2} + \frac{1}{a^2+b^2} \geq \frac{9}{2(a^2+b^2+c^2)}\]So far, we´ve got the relation:\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\]The denominator $a^2+b^2+c^2$ obeys the inequality (Chebyschev once again):\[3(a^2+b^2+c^2) \geq (a+b+c)(a+b+c)=1 \Rightarrow \frac{1}{a^2+b^2+c^2}\leq 3\]
\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\leq \frac{27}{2}\]
Thus, we have:\[\frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \geq \frac{1}{2}\]

 

[anemone]

Very well done, lfdahl, and thanks for participating!(Cool)

 

[Albert1]

My solution:

Spoiler

WLOG we can let $a \ge b \ge c$, and we get:

\[\frac{1}{b^2+c^2} \geq \frac{1}{a^2+c^2}\geq \frac{1}{a^2+b^2}\]Repeated use of Chebyschevs Sum Inequality:\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq 9\underbrace{(a+b+c)}_{=1} \left ( \frac{a^2}{b^2+c^2} + \frac{b^2}{a^2+c^2} + \frac{c^2}{a^2+b^2} \right )\\\\ 9 \left ( \frac{a^2}{b^2+c^2} + \frac{b^2}{a^2+c^2} + \frac{c^2}{a^2+b^2} \right )\geq 3\underbrace{(a+b+c)}_{=1} \left ( \frac{a}{b^2+c^2} + \frac{b}{a^2+c^2} + \frac{c}{a^2+b^2} \right ) \\\\ 3 \left ( \frac{a}{b^2+c^2} + \frac{b}{a^2+c^2} + \frac{c}{a^2+b^2} \right )\geq \underbrace{(a+b+c)}_{=1} \left ( \frac{1}{b^2+c^2} + \frac{1}{a^2+c^2} + \frac{1}{a^2+b^2} \right )\]Using the Arithmetic Harmonic Mean Inequality:\[\frac{1}{b^2+c^2} + \frac{1}{a^2+c^2} + \frac{1}{a^2+b^2} \geq \frac{9}{2(a^2+b^2+c^2)}\]So far, we´ve got the relation:\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\]The denominator $a^2+b^2+c^2$ obeys the inequality (Chebyschev once again):\[3(a^2+b^2+c^2) \geq (a+b+c)(a+b+c)=1 \Rightarrow \frac{1}{a^2+b^2+c^2}\leq 3\]
\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\leq \frac{27}{2}-----(*)\]
Thus, we have:\[\frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \geq \frac{1}{2}\]

Spoiler

\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\leq \frac{27}{2}-----(*)\]
Thus, we have:
\[\frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \geq \frac{1}{2}\]
a question in (*)
if $x\geq y \geq z$ then $x\geq z$ there is no doubt
but if $x\geq y \leq z$ then $x\geq z$ is not always true

 

[Greg]

Let $a,\,b$ and $c$ be positive real numbers for which $a+ b + c = 1$.

Prove that \(\displaystyle \frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+a^2}+\frac{c^3}{a^2+b^2}\ge \frac{1}{2}.\)

[sp]Why not use the AM-GM inequality?

$$a+b+c=1,\quad a,b,c>0 \\\Leftrightarrow\dfrac{a^3}{b^2+c^2}=\dfrac{b^3}{c^2+a^2}=\dfrac{c^3}{a^2+b^2}\Rightarrow a=b=c=\dfrac13$$

$$f(a,b,c)=\dfrac{a^3}{b^2+c^2}+\dfrac{b^3}{c^2+a^2}+\dfrac{c^3}{a^2+b^2}$$

$$\min(f(a,b,c))=3\sqrt[3]{\dfrac{\left(\dfrac13\right)^9}{8\left(\dfrac13\right)^6}}=\dfrac12$$[/sp]

 

[lfdahl]

Spoiler

\[27 \left ( \frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \right )\geq \frac{9}{2(a^2+b^2+c^2)}\leq \frac{27}{2}-----(*)\]
Thus, we have:
\[\frac{a^3}{b^2+c^2} + \frac{b^3}{a^2+c^2} + \frac{c^3}{a^2+b^2} \geq \frac{1}{2}\]
a question in (*)
if $x\geq y \geq z$ then $x\geq z$ there is no doubt
but if $x\geq y \leq z$ then $x\geq z$ is not always true

Hi, Albert!

Spoiler

You´re absolutely right: The implication: $x \ge\ y \le z \Rightarrow x\ge z $ is not always true!
Therefore, my solution is not "bullet proof". I´m sorry for having delivered a dubious answer

 

[Albert1]

Hi, Albert!

Spoiler

You´re absolutely right: The implication: $x \ge\ y \ge z \Rightarrow x\ge z $ is not always true!
Therefore, my solution is not "bullet proof". I´m sorry for having delivered a dubious answer

Spoiler

The implication: $x \ge\ y \ge z \Rightarrow x\ge z $ is not always true!

it should be :
$x \ge\ y \le z \Rightarrow x\ge z $ is not always true!

 

[lfdahl]

Spoiler

The implication: $x \ge\ y \ge z \Rightarrow x\ge z $ is not always true!

it should be :
$x \ge\ y \le z \Rightarrow x\ge z $ is not always true!

Spoiler

You´re right again