Prove the sum of a² and d² is equal the sum of b² and c².

[anemone]

Let $a,\,b,\,c,\,d\in \mathbb{R}$ with condition that $a+b\sqrt{2}+c\sqrt{3}+2d\ge \sqrt{10(a^2+b^2+c^2+d^2)}$.

Prove that $a^2+d^2=b^2+c^2$.

 

[anemone]

Let $a,\,b,\,c,\,d\in \mathbb{R}$ with condition that $a+b\sqrt{2}+c\sqrt{3}+2d\ge \sqrt{10(a^2+b^2+c^2+d^2)}$.

Prove that $a^2+d^2=b^2+c^2$.

Hint:

Spoiler

Cauchy-Schwarz Inequality

 

[kaliprasad]

Hint:

Spoiler

Cauchy-Schwarz Inequality

Spoiler

by Cauchy-Schawartz in equality we have

$(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

Or $10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

Or $\sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)$

from given condition and above we have
$\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

when

$\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)$
So $a = k, b^2= 2k^2,c^2=3k^2,d=2k$ and hence $a^2+d^2=b^2+c^2$

 

[anemone]

Spoiler

by Cauchy-Schawartz in equality we have

$(a^2+b^2+c^2+d^2)(1^2 + \sqrt2^2+ \sqrt3^2+2^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

Or $10(a^2 + b^2 + c^2+ d^2)>=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

Or $\sqrt{10(a^2 + b^2+ c^2+d^2)}>=(a+b\sqrt{2}+c\sqrt{3}+2d)$

from given condition and above we have
$\sqrt{10(a^2 + b^2+c^2+d^2)}=(a+b\sqrt{2}+c\sqrt{3}+2d)^2$

when

$\frac{a}{1}= \frac{b}{\sqrt2}=\frac{c}{\sqrt3}= \frac{d}{2}= k (say)$
So $a = k, b^2= 2k^2,c^2=3k^2,d=2k$ and hence $a^2+d^2=b^2+c^2$

Thanks kaliprasad for participating and the nice solution, I solved it a bit different than yours:

Spoiler

By applying the Cauchy Schwarz inequality to the LHS of the given inequality, we get:

$a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$

From $\sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}\le a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$, we can conclude that

$a+b\sqrt{2}+c\sqrt{3}+2d=\sqrt{10(a^2+b^2+c^2+d^2)}$.

Using Cauchy Schwarz inequality again, separately, on both $a+2d$ and $b\sqrt{2}+c\sqrt{3}$, we get:

$a+2d≤\sqrt{5(a^2+d^2)}$ and $b\sqrt{2}+c\sqrt{3}\le \sqrt{5(b^2+c^2)}$.

Adding them up and replacing $a+b\sqrt{2}+c\sqrt{3}+2d$ by $\sqrt{10(a^2+b^2+c^2+d^2)}$, we end up getting:

$\sqrt{a^2+d^2}+\sqrt{b^2+c^2}\le 0$, which is true iff $a^2+d^2=b^2+c^2$. (Q. E. D.)

 

[kaliprasad]

Thanks kaliprasad for participating and the nice solution, I solved it a bit different than yours:

Spoiler

By applying the Cauchy Schwarz inequality to the LHS of the given inequality, we get:

$a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$

From $\sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}\le a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$, we can conclude that

$a+b\sqrt{2}+c\sqrt{3}+2d=\sqrt{10(a^2+b^2+c^2+d^2)}$.

Using Cauchy Schwarz inequality again, separately, on both $a+2d$ and $b\sqrt{2}+c\sqrt{3}$, we get:

$a+2d≤\sqrt{5(a^2+d^2)}$ and $b\sqrt{2}+c\sqrt{3}\le \sqrt{5(b^2+c^2)}$.

Adding them up and replacing $a+b\sqrt{2}+c\sqrt{3}+2d$ by $\sqrt{10(a^2+b^2+c^2+d^2)}$, we end up getting:

$\sqrt{a^2+d^2}+\sqrt{b^2+c^2}\le 0$, which is true iff $a^2+d^2=b^2+c^2$. (Q. E. D.)

Spoiler

last line is in error as $\sqrt{a^2+d^2}+\sqrt{b^2+c^2}\le 0$ => a=b=c=d = 0 but for other values also it is true

 

[anemone]

Spoiler

last line is in error as $\sqrt{a^2+d^2}+\sqrt{b^2+c^2}\le 0$ => a=b=c=d = 0 but for other values also it is true

Argh...sorry kaliprasad, last night I wasn't feeling okay and therefore the solution that I posted (with typo) in a hurry didn't quite hold water.

I therefore want to give the reasoning that I've "in my mind" that would definitely work, to amend the incomplete solution I posted yesterday:

Spoiler

By applying the Cauchy Schwarz inequality to the LHS of the given inequality, we get:

$a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$

From $\sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}\le a+b\sqrt{2}+c\sqrt{3}+2d\le \sqrt{10}\sqrt{(a^2+b^2+c^2+d^2)}$, we can conclude that

$a+b\sqrt{2}+c\sqrt{3}+2d=\sqrt{10(a^2+b^2+c^2+d^2)}$.

Using Cauchy Schwarz inequality again, separately, on both $a+2d$ and $b\sqrt{2}+c\sqrt{3}$, we get:

$a+2d≤\sqrt{5(a^2+d^2)}$ and $b\sqrt{2}+c\sqrt{3}\le \sqrt{5(b^2+c^2)}$.

Adding them up and replacing $a+b\sqrt{2}+c\sqrt{3}+2d$ by $\sqrt{10(a^2+b^2+c^2+d^2)}$, we end up getting:

$\sqrt{10(a^2+b^2+c^2+d^2)}\le \sqrt{5(a^2+d^2)}+\sqrt{5(b^2+c^2)}$

$\sqrt{2(a^2+b^2+c^2+d^2)}\le \sqrt{a^2+d^2}+\sqrt{b^2+c^2}$

Squaring both sides gives

$2(a^2+b^2+c^2+d^2)\le a^2+d^2+2\sqrt{a^2+d^2}\sqrt{b^2+c^2}+b^2+c^2$

$a^2+d^2-2\sqrt{a^2+d^2}\sqrt{b^2+c^2}+b^2+c^2\le 0$

$(\sqrt{a^2+d^2}-\sqrt{b^2+c^2})^2\le 0$, which is true iff $a^2+d^2=b^2+c^2$. (Q. E. D.)