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

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The discussion centers on proving that for real numbers a, b, c, and d, the equation a² + d² = b² + c² holds under the condition a + b√2 + c√3 + 2d ≥ √10(a² + b² + c² + d²). Participants share their approaches to the proof, with one member acknowledging a previous error in their solution and expressing intent to clarify their reasoning. The conversation highlights the importance of rigorous proof and correction in mathematical discourse. The hint suggests that further exploration of the problem could lead to a valid solution. The thread emphasizes collaborative problem-solving in mathematics.
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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 $a^2+d^2=b^2+c^2$.
 
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anemone said:
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:

Cauchy-Schwarz Inequality
 
anemone said:
Hint:

Cauchy-Schwarz Inequality

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 $a^2+d^2=b^2+c^2$
 
kaliprasad said:
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:

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

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.)

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
 
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kaliprasad said:
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:

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.)
 
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