Show That These Equalities Hold

How do you prove the equality of polynomials, meaning for all polynomials?

Two polynomials [math]p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0[/math] and [math]q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0[/math] are equal if and only if [math]m=n[/math] and [math]a_k=b_k[/math] for all [math]k\in\{0,1,2,\ldots,n\}[/math].I can hear you say:“But, wait a second. If what you’re saying is true, then what does [math]x^2+x-6=0[/math] mean? By your definition, [math]x^2+x-6[/math] and [math]0[/math] are not equal polynomials… and yet here they are equal? This is confusing!”The answer to this question is that the equation [math]x^2+x-6=0[/math] is not saying that the two polynomials on either side of the equation are equal.We need the concept of the value of a polynomial at [math]c[/math] here. The value of the polynomial [math]p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0[/math] at [math]c[/math], denoted by [math]p(c)[/math], is [math]a_nc^n+a_{n-1}c^{n-1}+\cdots+a_1c+a_0[/math]. The equation [math]p(x)=0[/math] is then asking us to find every such [math]c[/math] for which the value of [math]p(x)[/math] at each of these [math]c[/math]’s is [math]0[/math]. These [math]c[/math]’s are called the roots of the polynomial [math]p(x)[/math].It is important to make this difference when dealing with polynomials, as it is quite subtle, notation-wise.

How do I show that [math]|z_1+z_2+z_3+\cdots+z_7|\leq |z_1|+|z_2|+|z_3|+\cdots+|z_7|[/math]?

[math]\def\Re{\textrm{Re}} \def\Im{\textrm{Im}}[/math] That’s the triangle inequality applied six times. Let’s show it once.I’m assuming from the [math]z[/math]s these are complex numbers. The inequality still holds.Let’s first show for complex [math]z=a+bi, w=c+di,[/math] with [math]a,b,c,d[/math] real,[math]|z|+|w| \ge |z+w| [/math]I’ll run the proof backwards from the way I derived it.[math](ad-bc)^2 \ge 0[/math][math]a^2 d^2 + b^2 c^2 \ge 2 abcd[/math][math] a^2 c^2 + b^2 d^2 + a^2 d^2 + b^2 c^2 \ge a^2 c^2 + b^2 d^2 + 2 abcd[/math][math](a^2+b^2)(c^2+d^2) \ge (ac+bd)^2[/math][math]|a+bi|^2 |c+di|^2 \ge (\Re((a+bi)(c-di)))^2[/math][math]|z|^2 |w|^2 \ge (\Re(zw^*))^2[/math]The magnitudes on the left are non-negative. The real part on the right side may or may not be. So we can drop the squares and maintain the inequality.[math]|z||w| \ge \Re(zw^*) [/math][math]|z||w| \ge \frac 1 2 ( zw^* + z^*w ) [/math][math] |z|^2 + |w|^2 + 2|z||w| \ge |z|^2 + |w|^2 + ( zw^* + z^*w) [/math][math] (|z| + |w|)^2 \ge zz^* + ww^* + ( zw^* + z^*w) [/math][math] (|z| + |w|)^2 \ge (z+w)(z^*+w^*) [/math][math] (|z| + |w|)^2 \ge (z+w)(z+w)^*[/math][math] (|z| + |w|)^2 \ge |z+w|^2[/math]Since the things being squared on both sides are non-negative, the inequality survives the square root:[math]|z| + |w| \ge |z+w| \quad\checkmark[/math]Turning to the long sum,[math]|z_1|+|z_2| \ge |z_1+z_2|[/math][math] |z_1|+|z_2| + |z_3| \ge |z_1+z_2| + |z_3| [/math][math] |z_1+z_2|+|z_3| \ge |z_1 + z_2 + z_3 | [/math][math] |z_1|+|z_2| + |z_3| \ge |z_1 + z_2 + z_3 | [/math]It’s easy to see how this extends to the sum of seven terms.