Prove $\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1$ with Algebra Challenge

[anemone]

Let $a,\,b,\,c$ and $d$ be positive real numbers such that $(a^2+b^2)^3=c^2+d^2$, prove that $\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1$.

 

[anemone]

Hint:

Spoiler

Cauchy-Schwarz Inequality

 

[anemone]

My solution:

Spoiler

Apply the Cauchy-Schwarz inequality to the sum of $a^2+b^2$, we have:

$$a^2+b^2=\sqrt{a}\sqrt{a^3}+\sqrt{b}\sqrt{b^3}=\frac{\sqrt{c}\sqrt{a}\sqrt{a^3}}{\sqrt{c}}+\frac{\sqrt{d}\sqrt{b}\sqrt{b^3}}{\sqrt{d}}\le \sqrt{ac+bd}\sqrt{\frac{a^3}{c}+\frac{b^3}{d}}$$

Now we rearrange it to make $$\frac{a^3}{c}+\frac{b^3}{d}$$ on the LHS of the inequality, we see that we get:

$$\frac{a^3}{c}+\frac{b^3}{d}\ge\frac{(a^2+b^2)^2}{ac+bd}$$(*)

Using again the Cauchy-Schwarz inequality to the sum of $ac+bd$ we have:

$$ac+bd\le\sqrt{a^2+b^2}\sqrt{c^2+d^2}$$

$$\begin{align*}\therefore \frac{a^3}{c}+\frac{b^3}{d}&\ge\frac{(a^2+b^2)^2}{ac+bd}\\&\ge \frac{(a^2+b^2)^2}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}}\\&\ge \frac{(a^2+b^2)^{\frac{3}{2}}}{(c^2+d^2)^{\frac{1}{2}}}\\&\ge \left(\frac{(a^2+b^2)^3}{(c^2+d^2)}\right)^{\frac{1}{2}}\\&\ge 1\,\,\,\,\,\,\,\,\text{(Q.E.D.)}\end{align*}$$

 

[Albert1]

Let $a,\,b,\,c$ and $d$ be positive real numbers such that $(a^2+b^2)^3=c^2+d^2---(1)$, prove that $\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1$.

my solution :

Spoiler

using $AP\geq GP$
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3d}{cd}=\dfrac{2a^3}{c}---(2)$
equality occurs when $a^3d=b^3c (\,\, that \,\ is :\ a=b\,\,and\,\, c=d)$
if so $(1)$ becomes:$8a^6=2c^2$ ,or $a^3=\dfrac {c}{2}$
and $(2)$ becomes:
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3}{c}=1$

 

[Greg]

Re: Albert's solution.

Spoiler

equality occurs when $a^3d=b^3c (\,\, that \,\ is :\ a=b\,\,and\,\, c=d)$

I don't think $a^3d=b^3c$ implies $a=b$ and $c=d$. Consider $2^3\cdot8$ and $4^3\cdot1$.

 

[Albert1]

Re: Albert's solution.

Spoiler

I don't think $a^3d=b^3c$ implies $a=b$ and $c=d$. Consider $2^3\cdot8$ and $4^3\cdot1$.

Spoiler

consider $a^2+b^2$ when $a=b$ minimun of $a^2+b^2$ occurs
also consider $c^2+d^2$ when $c=d$ minimun of $c^2+d^2$ occurs
and more if $a=2,b=4,c=1,d=8$
$(a^2+b^2)^3=c^2+d^2$ does not fit

 

[anemone]

My solution:

Spoiler

Apply the Cauchy-Schwarz inequality to the sum of $a^2+b^2$, we have:

$$a^2+b^2=\sqrt{a}\sqrt{a^3}+\sqrt{b}\sqrt{b^3}=\frac{\sqrt{c}\sqrt{a}\sqrt{a^3}}{\sqrt{c}}+\frac{\sqrt{d}\sqrt{b}\sqrt{b^3}}{\sqrt{d}}\le \sqrt{ac+bd}\sqrt{\frac{a^3}{c}+\frac{b^3}{d}}$$

Now we rearrange it to make $$\frac{a^3}{c}+\frac{b^3}{d}$$ on the LHS of the inequality, we see that we get:

$$\frac{a^3}{c}+\frac{b^3}{d}\ge\frac{(a^2+b^2)^2}{ac+bd}$$(*)

Using again the Cauchy-Schwarz inequality to the sum of $ac+bd$ we have:

$$ac+bd\le\sqrt{a^2+b^2}\sqrt{c^2+d^2}$$

$$\begin{align*}\therefore \frac{a^3}{c}+\frac{b^3}{d}&\ge\frac{(a^2+b^2)^2}{ac+bd}\\&\ge \frac{(a^2+b^2)^2}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}}\\&\ge \frac{(a^2+b^2)^{\frac{3}{2}}}{(c^2+d^2)^{\frac{1}{2}}}\\&\ge \left(\frac{(a^2+b^2)^3}{(c^2+d^2)}\right)^{\frac{1}{2}}\\&\ge 1\,\,\,\,\,\,\,\,\text{(Q.E.D.)}\end{align*}$$

I didn't know where my head was the time I composed my solution as I didn't include the condition when the equality occurs...sorry about it!

Spoiler

In my solution, equality occurs when $$\frac{a}{c}=\frac{b}{d}$$.

 

[anemone]

my solution :

Spoiler

using $AP\geq GP$
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3d}{cd}=\dfrac{2a^3}{c}---(2)$
equality occurs when $a^3d=b^3c (\,\, that \,\ is :\ a=b\,\,and\,\, c=d)$
if so $(1)$ becomes:$8a^6=2c^2$ ,or $a^3=\dfrac {c}{2}$
and $(2)$ becomes:
$\dfrac{a^3}{c}+\dfrac{b^3}{d}=\dfrac{a^3d+b^3c}{cd}\geq \dfrac{2a^3}{c}=1$

Spoiler

Hi Albert,

Albeit it's true that equality occurs when $a=b,\,c=d$ but I don't see how we could provide the airtight proof from $a^3d=b^3c$ and $(a^2+b^2)^3=c^2+d^2$ with the conclusion that $a=b$ and $c=d$...

 

[Albert1]

Spoiler

Hi Albert,

Albeit it's true that equality occurs when $a=b,\,c=d$ but I don't see how we could provide the airtight proof from $a^3d=b^3c$ and $(a^2+b^2)^3=c^2+d^2$ with the conclusion that $a=b$ and $c=d$...

Spoiler

$if \,a=b,\,\, then \\
\dfrac{8a^6}{2c^2}=\dfrac{c^2+d^2}{2c^2}\\
\therefore \dfrac{2a^3}{c}=\sqrt{\dfrac{c^2+d^2}{2c^2}}\geq\sqrt{\dfrac{2c
^2}{2c^2}}=1$
equality occurs when $c=d$