a) Cho hai số thực dương \[x;y\] thỏa mãn: x+ y bé hơn bằng 2/3

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a) Cho hai số thực dương \[x;y\] thỏa mãn: \[x + y\,\, \le \,\,\frac{2}{3}\]. Tìm giá trị nhỏ nhất của

\[A = 53x + 53y\,\, + \,\,\frac{1}{{{x^2}}}\, + \,\,\frac{1}{{{y^2}}}\].

b) Cho ba số thực dương \[x;y\,,\,z\] thỏa mãn: \[{x^2} + {y^2}\,\, + {z^2} \ge \,\,3\]. Chứng minh rằng:

\[\left( {{x^4} + {y^4}\,\, + {z^4}} \right)\,\, + \,\,\left( {{x^3} + {y^3}\,\, + {z^3}} \right) \ge \,\,3\, + x + y + z\].

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Giải thích

a) Dự đoán điểm rơi:

\[x = y\,\, = \,\,\frac{1}{3}\,\, \Rightarrow \,\frac{1}{{{x^2}}} + ax\, + ax\,\,\mathop \ge \limits^{Co - Si} \,\,3.\sqrt[3]{{\frac{1}{{{x^2}}} \cdot ax \cdot ax}} = 3.\sqrt[3]{{{a^2}}}\,\, \Rightarrow \,\,\frac{1}{{{x^2}}} = ax\,\, \Rightarrow a = 27\]

Ta có: \[A = 53x + 53y\,\, + \,\,\frac{1}{{{x^2}}}\, + \,\,\frac{1}{{{y^2}}}\, = \left( {\,27x + 27x\,\, + \,\,\frac{1}{{{x^2}}}} \right)\, + \,\,\left( {\,27y + 27y\,\, + \,\,\frac{1}{{{y^2}}}} \right) - \left( {x + y} \right)\]

\[ \Rightarrow A\,\,\,\mathop \ge \limits^{Co - Si} \,\,3.\sqrt[3]{{27x \cdot 27x\,\, \cdot \,\,\frac{1}{{{x^2}}}}}\, + \,\,3.\sqrt[3]{{27y \cdot 27y\,\, \cdot \,\,\frac{1}{{{y^2}}}}} - \left( {x + y} \right)\,\, = 27 + 27 - \,\left( {x + y} \right)\,\, \ge \,\,54 - \frac{2}{3} = \frac{{160}}{3}\]Dấu “=” xảy ra khi \[x = y\,\, = \,\,\frac{1}{3}\]

Vậy \[ \Rightarrow Min\,A\,\,\, = \frac{{160}}{3}\,\, \Leftrightarrow \,\,x = y = \frac{1}{3}\]

b) Ta có: \[{x^4}\, + \,1\,\, \ge \,2.\sqrt {{x^4}\,.\,1} = 2{x^2}\,\,;\,\,{y^4}\, + \,1\,\, \ge \,2.\sqrt {{y^4}\,.\,1} = 2{y^2}\,\,;\,\,{z^4}\, + \,1\,\, \ge \,2.\sqrt {{z^4}\,.\,1} = 2{z^2}\]

\[ \Rightarrow {x^4}\, + \,{y^4} + {z^4}\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right) - 3\,\, \Rightarrow \,\,VT\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right) - 3\,\, + \,{x^3} + {y^3} + {z^3}\]

Tương tự: \[{x^3}\, + \,x\,\, \ge \,2.\sqrt {{x^3}\,.\,x} = 2{x^2}\,\,;\,\,{y^3}\, + \,y\,\, \ge \,2.\sqrt {{y^3}\,.\,y} = 2{y^2}\,\,;\,\,{z^3}\, + \,z\,\, \ge \,2.\sqrt {{z^3}\,.\,z} = 2{z^2}\]

\[\begin{array}{l} \Rightarrow {x^3}\, + \,{y^3}\, + {z^3}\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\,\,\\ \Rightarrow \,\,\,VT\,\, \ge \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,2\left( {{x^2} + {y^2} + {z^2}} \right)\, - 3\end{array}\]

\[\begin{array}{l} \Rightarrow \,\,\,VT\,\, \ge \,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,3\left( {{x^2} + {y^2} + {z^2}} \right)\, - 3\,\\ \Rightarrow VT\, \ge \,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,3.3\, - 3\end{array}\]\[ \Rightarrow \,\,\,VT\,\, \ge \,\,\,\left( {{x^2} + {y^2} + {z^2}} \right)\, - \left( {x + y + z} \right)\,\, + \,\,6\]

Mà: \[{x^2}\, + \,1\,\, \ge \,2.\sqrt {{x^2}\,.\,1} = 2x\,\,;\,\,{y^2}\, + \,1\,\, \ge \,2.\sqrt {{y^2}\,.\,1} = 2y\,\,;\,\,{z^2}\, + \,1\,\, \ge \,2.\sqrt {{z^2}\,.\,1} = 2z\]

\[ \Rightarrow {x^2}\, + \,{y^2} + {z^2}\,\, \ge \,\,2\left( {x + y + z} \right) - 3\,\,\,\]

\[ \Rightarrow \,\,VT\,\, \ge \,\,2\left( {x + y + z} \right) - 3\, - \,\left( {x + y + z} \right) + 6\, = \,\left( {x + y + z} \right) + 3\] (đpcm)