Let $a,b,c$ be positive numbers such that $a^{2}+b^{2}-ab=c^{2}$.prove that,
$(a-c)(b-c)=<0$.
INEQUALITY
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- Phlembac Adib Hasan
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Re: INEQUALITY
Given that, \[a^2+b^2-ab=c^2 \]
WLOG we can take $a\ge b$.Surely $a \ge c $ [Otherwise it isn't possible.]
so we can say,\[(a-b)^2=c^2-ab\]
\[\Rightarrow c^2 \ge ab \ge b^2 \]
\[\Rightarrow c \ge b \]
Which completes our proof.
Comment:You should post it in Secondary level.
WLOG we can take $a\ge b$.Surely $a \ge c $ [Otherwise it isn't possible.]
so we can say,\[(a-b)^2=c^2-ab\]
\[\Rightarrow c^2 \ge ab \ge b^2 \]
\[\Rightarrow c \ge b \]
Which completes our proof.
Comment:You should post it in Secondary level.
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Re: INEQUALITY
It's easy to prove a,b,c forms a triangle by co-sine law & $\angle C=60\circ $ .
case 1:Then, if ABC is equilateral (a−c)(b−c)=0 .
or,case-2: $\angle A>60\circ $ then, a-c>0;b-c<0; proved.
or,case-3: $\angle B>60\circ $ ,b>a. a-b<0;b-c>0 ,so,proved.
case 1:Then, if ABC is equilateral (a−c)(b−c)=0 .
or,case-2: $\angle A>60\circ $ then, a-c>0;b-c<0; proved.
or,case-3: $\angle B>60\circ $ ,b>a. a-b<0;b-c>0 ,so,proved.