Question

Prove that √3 + √7 < 2√5

Answer 1

To solve these kind of questions, the key rule is $\blue{squaring on both sides}$

so

${( \sqrt{3} + \sqrt{7} )}^{2}$

we know

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

so

${ \sqrt{3} }^{2} + 2( \sqrt{3} )( \sqrt{7} ) + { \sqrt{7} }^{2}$

$3 + 2 \sqrt{21} + 7$

$10 + 2 \sqrt{21}$

now let's take R.H.S

${(2 \sqrt{5}) }^{2}$

$4 \times { \sqrt{5} }^{2}$

$4 \times 5$

$20$

so L.H.S sq is 10+2√21 which is approx 10+2(4.5) so nearly close to 19..

but R.H.S sq is 20

so L.H.S sq is less than R.H.S. sq

hence proved...!!!