Question Number 141651 by mathsuji last updated on 21/May/21 | ||
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Commented by MJS_new last updated on 22/May/21 | ||
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$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$$${a}={b}={c}=\mathrm{0}\:\Rightarrow\:\mathrm{6}\sqrt{\mathrm{2}}>\mathrm{0} \\ $$$${a}={b}={c}=−\sqrt{\mathrm{3}}\:\Rightarrow\:\mathrm{6}\sqrt{\mathrm{2}}>−\mathrm{6}\sqrt{\mathrm{3}}>−\mathrm{9}\sqrt{\mathrm{3}} \\ $$ | ||
Commented by mathsuji last updated on 22/May/21 | ||
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$${sorry}\:{sir},\:{a},{b},{c}>\mathrm{0} \\ $$ | ||