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[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex]\rm \longmapsto\:a > 0 \: \: and \: \: b > 0[/tex]
and
[tex]\rm \longmapsto\: {a}^{6} + {b}^{6} \leqslant 12 {a}^{2} {b}^{2} - 64[/tex]
can be rewritten as
[tex]\rm \longmapsto\: {a}^{6} + {b}^{6} + 64 - 12 {a}^{2} {b}^{2} \leqslant 0[/tex]
[tex]\rm \longmapsto\: {( {a}^{2}) }^{3} + {( {b}^{2} )}^{3} + {(4)}^{3} - 3( {a}^{2})( {b}^{2})(4) \leqslant 0[/tex]
We know
[tex]\mathfrak{ {x}^{3}+{y}^{3}+{z}^{3}-3xyz =(x + y + z)( {x}^{2}+{y}^{2}+{z}^{2}-xy-yz - zx}[/tex]
So, using this identity, we get
[tex] \rm( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) \leqslant 0[/tex]
Let we consider,
[tex]\bf{ \longmapsto\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}}[/tex]
can be rewritten as
[tex]\bf{ \: = \dfrac{1}{2}\bigg[2{a}^{4} + 2{b}^{4} + 32 - 2{a}^{2} {b}^{2} - 8 {b}^{2} - 8{a}^{2}\bigg]}[/tex]
[tex]\bf{ = \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]}[/tex]
can be re-arranged as
[tex]\bf{ \: = \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]}[/tex]
[tex]\bf{ \: = \dfrac{1}{2}\bigg[( {a}^{2} - {b}^{2})^{2} + {( {b}^{2} - 4) }^{2} + {( {a}^{2} - 4)}^{2} \bigg]}[/tex]
As sum of squares can never be negative.
[tex]\bf {⇛\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2} \geqslant 0}[/tex]
[ Equality of zero holds when a = b = 2 ]
And if a and b are distinct, then
[tex]\bf{⇛\:{a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2} > 0}[/tex]
and
[tex]\bf{( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) > 0}[/tex]
So,
[tex] \bf{( {a}^{2}+{b}^{2} + 4)( {a}^{4} + {b}^{4} + 16 - {a}^{2} {b}^{2} - 4 {b}^{2} - 4 {a}^{2}) \geqslant 0}[/tex]
So, for the given statement,
[tex]\rm \longmapsto\: {a}^{6} + {b}^{6} \geqslant 12 {a}^{2} {b}^{2} - 64[/tex]
So, we concluded that
If a = b = 2, then
[tex]\rm \longmapsto\: {a}^{6} + {b}^{6} = 12 {a}^{2} {b}^{2} - 64[/tex]
And
If a and b are distinct, then
[tex]\rm \longmapsto\: {a}^{6} + {b}^{6} > 12 {a}^{2} {b}^{2} - 64[/tex]
So, given statement is partially true for a = b = 2 otherwise false.