Question

IF a b c are positive real no.'s State if this is True or False
Also, Give a Proper Explanation and show the working.
I will mark the Best Answer Brainliest

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:a > 0 \: \: and \: \: b > 0$

and

$\rm :\longmapsto\: {a}^{6} + {b}^{6} \leqslant 12 {a}^{2} {b}^{2} - 64$

can be rewritten as

$\rm :\longmapsto\: {a}^{6} + {b}^{6} + 64 - 12 {a}^{2} {b}^{2} \leqslant 0$

$\rm :\longmapsto\: {( {a}^{2}) }^{3} + {( {b}^{2} )}^{3} + {(4)}^{3} - 3( {a}^{2})( {b}^{2})(4) \leqslant 0$

We know

$\boxed{\tt{ {x}^{3}+{y}^{3}+{z}^{3}-3xyz =(x + y + z)( {x}^{2}+{y}^{2}+{z}^{2}-xy-yz - zx}}$

So, using this identity, we get

$\rm( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) \leqslant 0$

Let we consider,

$\red{\rm :\longmapsto\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}}$

can be rewritten as

$\red{\rm \:  = \dfrac{1}{2}\bigg[2{a}^{4} + 2{b}^{4} + 32 - 2{a}^{2} {b}^{2} - 8 {b}^{2} - 8{a}^{2}\bigg]}$

$\red{\rm \:  = \dfrac{1}{2}\bigg[{a}^{4} + {a}^{4} + {b}^{4} + {b}^{4} + 16 + 16 - 2{a}^{2} {b}^{2} - 8 {b}^{2} - 8{a}^{2}\bigg]}$

can be re-arranged as

$\red{\rm \:  = \dfrac{1}{2}\bigg[({a}^{4} + {b}^{4} - 2{a}^{2} {b}^{2})+({b}^{4} + 16- 8 {b}^{2}) + (16 + {a}^{4} - 8{a}^{2})\bigg]}$

$\red{\rm \:  = \dfrac{1}{2}\bigg[( {a}^{2} - {b}^{2})^{2} + {( {b}^{2} - 4) }^{2} + {( {a}^{2} - 4)}^{2} \bigg]}$

As sum of squares can never be negative.

$\red{\rm \implies\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2} \geqslant 0}$

[ Equality of zero holds when a = b = 2 ]

And if a and b are distinct, then

$\red{\rm \implies\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2} > 0}$

and

$\red{\rm( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) > 0}$

So,

$\rm( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) \geqslant 0$

So, for the given statement,

$\rm :\longmapsto\: {a}^{6} + {b}^{6} \geqslant 12 {a}^{2} {b}^{2} - 64$

So, we concluded that

If a = b = 2, then

$\rm :\longmapsto\: {a}^{6} + {b}^{6} = 12 {a}^{2} {b}^{2} - 64$

And

If a and b are distinct, then

$\rm :\longmapsto\: {a}^{6} + {b}^{6} > 12 {a}^{2} {b}^{2} - 64$

So, given statement is partially true for a = b = 2 otherwise false.