Toán Lớp 8: Chứng minh : a^2+b^2+1≥ab +a+b
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Toán Lớp 8: Chứng minh : a^2+b^2+1≥ab +a+b
Comments ( 1 )
{\left( {a – b} \right)^2} \ge 0,\,\,\,\forall a,b\\
\Leftrightarrow {a^2} – 2ab + {b^2} \ge 0,\,\,\,\forall a,b\\
\Leftrightarrow {a^2} + {b^2} \ge 2ab,\,\,\,\forall a,b\\
{\left( {a – 1} \right)^2} \ge 0,\,\,\,\forall a\\
\Leftrightarrow {a^2} – 2.a.1 + {1^2} \ge 0,\,\,\,\forall a\\
\Leftrightarrow {a^2} – 2a + 1 \ge 0,\,\,\,\forall a\\
\Leftrightarrow {a^2} + 1 \ge 2a,\,\,\,\forall \,a\\
{\left( {b – 1} \right)^2} \ge 0,\,\,\,\forall b\\
\Leftrightarrow {b^2} – 2.b.1 + {1^2} \ge 0,\,\,\,\forall b\\
\Leftrightarrow {b^2} – 2b + 1 \ge 0,\,\,\,\forall b\\
\Leftrightarrow {b^2} + 1 \ge 2b,\,\,\,\forall b\\
\Rightarrow \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + 1} \right) + \left( {{b^2} + 1} \right) \ge 2ab + 2a + 2b,\,\,\,\forall \,a,b\\
\Leftrightarrow 2\left( {{a^2} + {b^2} + 1} \right) \ge 2\left( {ab + a + b} \right),\,\,\,\forall \,a,b\\
\Leftrightarrow {a^2} + {b^2} + 1 \ge ab + a + b,\,\,\,\forall \,a,b
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