[Toán 10]c/m bdt

E

eye_smile

a3a2+b2=a−ab2a2+b2≥a−ab22ab=a−b2\dfrac{a^3}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2} \ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}

b3b2+c2≥b−c2\dfrac{b^3}{b^2+c^2} \ge b-\dfrac{c}{2}

c3c2+a2≥c−a2\dfrac{c^3}{c^2+a^2} \ge c-\dfrac{a}{2}

Cộng theo vế \Rightarrow đpcm
 
You must log in or register to reply here.
Top Bottom