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Question Number 197848 by cortano12 last updated on 01/Oct/23

Answered by mr W last updated on 01/Oct/23

$$f\left(x\right)=\frac{a^2}{{\cos }^{2}x}+\frac{b^2}{{\sin }^{2}x}⩾\frac{{(a+b)}^2}{{\cos }^{2}x+{\sin }^{2}x}={(a+b)}^2$$$$\Rightarrow f{\left(x\right)}_{min}={(a+b)}^2$$

Commented by cortano12 last updated on 01/Oct/23

$$\mathrm{Titus}\mathrm{Lemma}\mathrm{sir}?$$

Commented by mr W last updated on 01/Oct/23

Answered by MM42 last updated on 01/Oct/23

$$f\left(x\right)=a^2+b^2+b^2{tan}^{2}x+a^2{cot}^{2}x$$$$f^{{}^{\prime }}\left(x\right)=2b^{2}tanx(1+{tan}^{2}x)-2a^{2}cotx(1+{cot}^{2}x)$$$$f^{\prime }\left(x\right)=0\Rightarrow {tan}^{2}x=\frac{a}{b}$$$$\Rightarrow minf=a^2+b^2+b^2\times \frac{a}{b}+a^2\times \frac{b}{a}$$$$={(a+b)}^2✓$$

Answered by dimentri last updated on 01/Oct/23

$$f\left(x\right)=\frac{a^2\sin {}^{2}x}{\cos {}^{2}x}+\frac{b^2\cos {}^{2}x}{\sin {}^{2}x}+a^2+b^2$$$$f\left(x\right)=a^2\tan {}^{2}x+b^2\cot {}^{2}x+a^2+b^2$$$$f\left(x\right)⩾2\sqrt{a^2\tan {}^{2}x.b^2\cot {}^{2}x}+a^2+b^2$$$$f\left(x\right)⩾2ab+a^2+b^2$$$$f\left(x\right)⩾{(a+b)}^2$$$$minf\left(x\right)={(a+b)}^2$$

Answered by mnjuly1970 last updated on 01/Oct/23

$$f\left(x\right)=a^2+b^2+a^2{tan}^2\left(x\right)+b^2{cot}^2\left(x\right)$$$$-----$$$$a^2{tan}^2\left(x\right)+b^2{cot}^2\left(x\right)⩾2\sqrt{a^2{b}^2}$$$$--$$$$f\left(x\right)\stackrel{a,b>0}{⩾}{a}^2+b^2+2ab={(a+b)}^2$$$${min}_f={(a+b)}^2$$

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