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Relevant wiki: Power Mean Inequality (QAGH)

According to Power-Mean Inequality , the Quadratic Mean is more than or equal to the Arithmetic Mean of the data values:

$\sqrt{\dfrac{a^2 + b^2 +c^2}{3}} \geq \dfrac{a+b+c}{3}$

$a^2 + b^2 +c^2 \geq \dfrac{(a+b+c)^2}{3}$

$\dfrac{3(a^2 + b^2 +c^2)}{a+b+c}\geq a+b+c$

From the question, $a<b<c$ , so such equality will not hold for this case, and since the rectangle's area equals that of the three squares, then we can set up:

$d(a+b+c) = a^2 + b^2 +c^2$ . Thus, $d = \dfrac{a^2 + b^2 +c^2}{a+b+c}$ .

As a result, $3d = d+d+d > a+b+c$ .

$d(a + b + c) = a^{2} + b^{2} + c^{2}$

Using the AM-GM inequality, equality cannot hold because $a \neq b \neq c$

$d(a + b + c) > 2ab + c^{2}$

$> 2bc + a^{2}$

$> 2ca + b^{2}$

Adding these together:

$3d(a + b + c) > a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = (a + b + c)^{2}$

$3d = d + d + d > a + b + c$

We have that:

$d(a+b+c) = a^2 + b^2 +c^2$ $\Rightarrow d = \frac{a^2 + b^2 +c^2}{a+b+c}$

By Titu's Lemma:

$d = \frac{a^2 + b^2 +c^2}{a+b+c} = \frac{a^2}{a+b+c} + \frac{b^2}{a+b+c} + \frac{c^2}{a+b+c}$

$\ge \frac{(a+b+c)^2}{ 3(a+b+c) }$

$\Rightarrow d \ge \frac{(a+b+c)}{ 3 } \Leftrightarrow 3d \ge a+b+c$

Equality holds when $a=b=c=d$ , but we know that $a \lt b \lt c$ . Hence the equality case must not hold.

Therefore, it is the case that:

$\large \boxed{ d + d + d \gt a+b+c}$

$a^2 + b^2 + c^2 = d(a+b+c) \\ \implies (a^2 + b^2 + c^2)(3) \ge (a+b+c)^2 \\ \implies 3d(a+b+c) \ge (a+b+c)^2 \\ \boxed {3d = d + d + d \ge a+b+c} .$