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[Dinhxuanbaohung]

Cho a,b,c > 0 thỏa mãn $a^{2} + b^{2}= c^{2} +1$

Tìm min 

P = $\sqrt{\frac{a^{3}}{a^{3} + (b+c)^3}} +\sqrt{\frac{b^{3}}{b^{3} + (a+c)^3}} + \frac{2c^3 + 1}{27}$

[NguyThang khtn]

Cho a,b,c > 0 thỏa mãn $a^{2} + b^{2}= c^{2} +1$

Tìm min 

P = $\sqrt{\frac{a^{3}}{a^{3} + (b+c)^3}} +\sqrt{\frac{b^{3}}{b^{3} + (a+c)^3}} + \frac{2c^3 + 1}{27}$

Ta có:

$\sqrt{\frac{a^{3}}{a^{3} + (b+c)^3}} $

$= \sqrt{\frac{1}{1+ \frac{(b+c)^3}{a^3}}} $

$= \sqrt{\frac{1}{(1+ \frac{b+c}{a})(1-\frac{b+c}{a}+\frac{(b+c)^2}{a^2})}}$

$\geq \frac{2}{2+\frac{(b+c)^2}{a^2}}$

$= \frac{2a^2}{2a^2+(b+c)^2}$

$ \geq \frac{2a^2}{2(a^2+b^2+c^2)}$

 

Tương tự ta cũng có : $\sqrt{\frac{b^{3}}{b^{3} + (a+c)^3}} \geq \frac{2b^2}{2(a^2+b^2+c^2)}$

Do đó :

$ P \geq \frac{2(a^2+b^2)}{2(a^2+b^2+c^2)}+\frac{2c^3 + 1}{27}$

$ = \frac{1+c^2}{1+2c^2}+\frac{2c^3 + 1}{27}$

 

Đến đây chắc khảo sát hàm này

Edited by NguyThang khtn, 15-01-2014 - 12:39.