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Question Number 175593 by Shrinava last updated on 03/Sep/22

If 0<a≤b then prove that: determinant ((1,a,(log(a^a ))),(1,((√((a^2 +b^2 )/2)) (√((a^2 +b^2 )/8)) ∙ log(((a^2 +b^2 )/2))),),(1,b,(log(b^b ))))≥ 0

$$\mathrm{If}\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}\:\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$ $$\begin{vmatrix}{\mathrm{1}}&{\mathrm{a}}&{\mathrm{log}\left(\mathrm{a}^{\boldsymbol{\mathrm{a}}} \right)}\\{\mathrm{1}}&{\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}}\:\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:\mathrm{log}\left(\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\right)}&{}\\{\mathrm{1}}&{\mathrm{b}}&{\mathrm{log}\left(\mathrm{b}^{\boldsymbol{\mathrm{b}}} \right)}\end{vmatrix}\geqslant\:\mathrm{0} \\ $$

Commented byRasheed.Sindhi last updated on 03/Sep/22

determinant ((1,( a),(log(a^a ))),(1,((√((a^2 +b^2 )/2))(√((a^2 +b^2 )/8)) ∙ log(((a^2 +b^2 )/2))),( ?)),(1,( b),(log(b^b ))))≥ 0

$$\begin{vmatrix}{\mathrm{1}}&{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}}&{\mathrm{log}\left(\mathrm{a}^{\boldsymbol{\mathrm{a}}} \right)}\\{\mathrm{1}}&{\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:\mathrm{log}\left(\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{2}}\right)}&{\:\:\:\:\:\:?}\\{\mathrm{1}}&{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}}&{\mathrm{log}\left(\mathrm{b}^{\boldsymbol{\mathrm{b}}} \right)}\end{vmatrix}\geqslant\:\mathrm{0} \\ $$ $$ \\ $$

Commented byJDamian last updated on 03/Sep/22

both square roots lie on the second column?

$${both}\:{square}\:{roots}\:{lie}\:{on}\:{the} \\ $$ $${second}\:{column}? \\ $$

Commented byShrinava last updated on 03/Sep/22

1 (√((a^2 +b^2 )/2)) (√((a^2 +b^2 )/8)) ∙ log(((a^2 +b^2 )/2))

$$\mathrm{1}\:\:\:\:\:\:\:\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}\:\:\:\:\:\:\:\:\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{8}}}\:\centerdot\:{log}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$

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