Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 55211 by naka3546 last updated on 19/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19

approach.. ((a^2 +(b−1)^2 )/2)≥(√(a^2 ×(b−1)^2 )) x^2 ≥2a(b−1) y^2 ≥2b(c−1) z^2 ≥2c(a−1) since x^2 =+ve so b>1 and a>0 for min value of x [ b=1] similarly for min value of y [c=1] for min value of z [a=1] x+y+z =(√(a^2 +(b−1)^2 )) +(√(b^2 +(c−1)^2 )) +(√(c^2 +(a−1)^2 )) =(√(1+0)) +(√(1+0)) +(√(1+0)) =3(min value)...

approach..a2+(b−1)22⩾a2×(b−1)2x2⩾2a(b−1)y2⩾2b(c−1)z2⩾2c(a−1)sincex2=+vesob>1anda>0forminvalueofx[b=1]similarlyforminvalueofy[c=1]forminvalueofz[a=1]x+y+z=a2+(b−1)2+b2+(c−1)2+c2+(a−1)2=1+0+1+0+1+0=3(minvalue)...

Commented by naka3546 last updated on 19/Feb/19

(√(a^2 + (b−1)^2 )) + (√(b^2 + (c−1)^2 )) + (√(c^2 + (a − 1)^2 )) ≥ 3 ? Is it right ?

a2+(b−1)2+b2+(c−1)2+c2+(a−1)2⩾3?Isitright?

Terms of Service

Privacy Policy

Contact: info@tinkutara.com