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Q.For any two vectors a⃗ and b⃗, prove that |a⃗ + b⃗| ≤ |a⃗| + |b⃗|. Also write the name of this inequality.

Punjab PsebPSEB Punjab Class 12 Board 2019Subjective· 4mImportance★★★★★
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Expand ∣a⃗+b⃗∣2|\vec a+\vec b|^2 as a dot product and bound the cross term using the Cauchy-Schwarz inequality.

We want to prove: for any two vectors a⃗,b⃗\vec a,\vec b,

∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec a+\vec b| \le |\vec a|+|\vec b|

Proof: Start with the square of the left side, using ∣v⃗∣2=v⃗⋅v⃗|\vec v|^2=\vec v\cdot\vec v:

∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)=a⃗⋅a⃗+2 a⃗⋅b⃗+b⃗⋅b⃗=∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2|\vec a+\vec b|^2 = (\vec a+\vec b)\cdot(\vec a+\vec b) = \vec a\cdot\vec a + 2\,\vec a\cdot\vec b + \vec b\cdot\vec b = |\vec a|^2+2\,\vec a\cdot\vec b+|\vec b|^2

Now, a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta where θ\theta is the angle between them, and since cos⁡θ≤1\cos\theta \le 1:

a⃗⋅b⃗≤∣a⃗∣∣b⃗∣\vec a\cdot\vec b \le |\vec a||\vec b|

Substituting this bound:

∣a⃗+b⃗∣2≤∣a⃗∣2+2∣a⃗∣∣b⃗∣+∣b⃗∣2=(∣a⃗∣+∣b⃗∣)2|\vec a+\vec b|^2 \le |\vec a|^2+2|\vec a||\vec b|+|\vec b|^2 = \big(|\vec a|+|\vec b|\big)^2

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