if such $a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$ show $2(a+b+c)=3$ Let $a,b,c,d\in R$ and such
$$a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$$
show that $$2(a+b+c)=3$$
It seem can use inequality to solve it?

Let $a+b+c-d=x^2$, where $x\geq0$.
Hence, $$a^2+b^2+c^2+1=a+b+c-x^2+x$$ or
$$x^2-x+a^2+b^2+c^2-a-b-c+1=0$$ or
$$\left(x-\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2=0,$$
which gives $a=b=c=\frac{1}{2}$ and $2(a+b+c)=3$.
Done!
