Cho xy + yz + zx 1. Chứng minh: x / (x^2 + 1) + y / (căn bậc hai (y^2 + 1)
Cho xy + yz + zx = 1. Chứng minh: \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{y}{{\sqrt {{y^2} + 1} }} + \frac{z}{{\sqrt {{z^2} + 1} }} \le \frac{3}{2}\).
Cho xy + yz + zx = 1. Chứng minh: \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{y}{{\sqrt {{y^2} + 1} }} + \frac{z}{{\sqrt {{z^2} + 1} }} \le \frac{3}{2}\).
Ta có: \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{y}{{\sqrt {{y^2} + 1} }} + \frac{z}{{\sqrt {{z^2} + 1} }}\)
\( = \frac{x}{{\sqrt {{x^2} + xy + yz + zx} }} + \frac{y}{{\sqrt {{y^2} + xy + yz + zx} }} + \frac{z}{{\sqrt {{z^2} + xy + yz + zx} }}\)
\( = \frac{x}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }} + \frac{y}{{\sqrt {\left( {y + z} \right)\left( {y + x} \right)} }} + \frac{z}{{\sqrt {\left( {z + x} \right)\left( {z + y} \right)} }}\)
Áp dụng BĐT Cô-si, ta có:
• \(\frac{x}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }} \le \frac{1}{2}\left( {\frac{x}{{x + y}} + \frac{x}{{x + z}}} \right)\)
• \(\frac{y}{{\sqrt {\left( {y + z} \right)\left( {y + x} \right)} }} \le \frac{1}{2}\left( {\frac{y}{{y + z}} + \frac{y}{{y + x}}} \right)\)
• \(\frac{z}{{\sqrt {\left( {z + x} \right)\left( {z + y} \right)} }} \le \frac{1}{2}\left( {\frac{z}{{z + x}} + \frac{z}{{z + y}}} \right)\)
Cộng vế theo vế:
\(\frac{x}{{\sqrt {\left( {x + y} \right)\left( {x + z} \right)} }} + \frac{y}{{\sqrt {\left( {y + z} \right)\left( {y + x} \right)} }} + \frac{z}{{\sqrt {\left( {z + x} \right)\left( {z + y} \right)} }}\)
\( \le \frac{1}{2}\left( {\frac{{x + y}}{{x + y}} + \frac{{y + z}}{{y + z}} + \frac{{z + x}}{{z + x}}} \right) = \frac{3}{2}\).
Dấu “=” xảy ra khi và chỉ khi \(x = y = z = \frac{1}{{\sqrt 3 }}\).
Vậy \(\frac{x}{{\sqrt {{x^2} + 1} }} + \frac{y}{{\sqrt {{y^2} + 1} }} + \frac{z}{{\sqrt {{z^2} + 1} }} \le \frac{3}{2}\) khi \(x = y = z = \frac{1}{{\sqrt 3 }}\).