$(a-2b)^2+(b-2c)^2 = 0 \implies a = 2b = 4c$. But, $a = 4c > 3c = c+2c = c+b$. Therefore, they can't form the sides of a triangle.

$a^2+4b^2+b^2+4c^2-4ab-4bc=0$ yielding $(a-2b)^2+(b-2c)^2=0$ yielding $a=4c,b=2c$. Thus $b+c=3c<4c=a$ concluding that a triangle cannot be formed.

$a^2 + 5b^2 + 4c^2 - 4ab-4bc = 0 \Rightarrow \left( a - 2b \right) ^ 2 + \left( b - 2c \right)^2 = 0$ $\Rightarrow a = 2b$ and $b = 2c$ $\Rightarrow a = 4c, b = 2c$ and $a = 4c > b + c = 3c$ $\Rightarrow a, b, c$ can not be the lengths of the sides of a triangle

$a^{2} + 5b^{2} + 4c^{2} - 4ab - 4bc = 0$ Hence we get $(a - 2b)^{2} + (b - 2c)^{2} = 0$ which implies that $a = 2b$ and $b = 2c$ which implies that $a=4c$. Thus we get that $b + c = 3c$ which is less than $4c$ since $4c$ is equal to $a$, $b + c < a$ which implies that $a, b, c$ cannot be the sides of a triangle.

Four times exactly the same solution .... Congrats all

it is not that hard to proove $ a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0 $ Simplifying we get, $ \implies (a-2b)^2 +(b-2c)^2 = 0$ $ \implies a=2b=4c$ if a, b, c were to be sides of a $ \Delta $ then, $ b+c>a $ $ \implies 2c+c > 4c $ contradiction Hence a,b,c cannot be the sides of a $ \Delta $

My Solutions. Solution 1. We can write the expression as follows $-$ $a^2+5b^2+4c^2-4ab-4bc=0\Rightarrow (a^2-4ab+4b^2)+(b^2-4bc+4c^2)=0\Rightarrow (a-2b)^2+(b-2c)^2=0.$ Now, $(a-2b)^2,(b-2c)^2\ge 0.$ And as they add up to $0,$ we can say that $a-2b=0=b-2c.$ Or, $b=2c,a=2b=4c.$ Then, $a=4c>3c=2c+c=b+c.$ If $a,b,c$ are sides of a triangle, then they must follow the triangle inequality, that is $a<b+c.$ But we are getting $a>b+c,$ which means the given real numbers $a,b,c$ cannot be sides of a triangle. Solution 2. By $AM-GM$ inequality, we have $a^2+4b^2\ge 2\sqrt{a^2\cdot 4b^2}=4ab\Rightarrow a^2+4b^2-4ab\ge 0.~~~(I)$ Similarly, $b^2+4c^2-4bc\ge 0.~~~(II)$ Adding up the inequalities we get $ a^2 + 5b^2 + 4c^2 - 4ab - 4bc \ge 0,$ but we know that $ a^2 + 5b^2 + 4c^2 - 4ab - 4bc = 0 $ This means, the equality cases will occur in $(I),(II).$ Means, $a^2=4b^2,b^2=4c^2\Rightarrow b=2c,a=2b=4c.$ From here, we can say that $a=4c>3c=2c+c=b+c.$ If $a,b,c$ are sides of a triangle, then they must follow the triangle inequality, that is $a<b+c.$ But we are getting $a>b+c,$ which means the given real numbers $a,b,c$ cannot be sides of a triangle.

Nice solution

too easy

Why are people writing the same solution again and again and it is too obvious that it is sum of 2 perfect squares

LoveMaths26102003 wrote: Why are people writing the same solution again and again and it is too obvious that it is sum of 2 perfect squares They want to show that they know how to apply $a^2-2ab+b^2=(a-b)^2$

Dear AOPSers, below I present my ability to write $(a-b)^2=a^2-2ab+b^2$. Notice that the given equation reduces to $(a-2b)^2+(b-2c)^2=0$. So, $a=2b=4c$. Now $b+c=\frac{3a}{4}<a$, contradicting the triangle inequality. Thus $a,b,c$ cannot form the sides of a triangle.

It simplifies to $(a-2b)^2 + (b-2c)^2 = 0$, so $(a, b, c) = (4k, 2k, 1k)$. But $2k + 1k < 4k$, contradicting triangle inequality. $\square$